JavaScript (ES8), 196 bytes

a=>`141

101

521

031

236

330

332`.replace(/./g,n=>[...`@-@   |@||||`.substr(n*3,3)].join(' -'[+!+n].repeat(m/2))||a[n-=4].padStart(m+l[n]+3>>1).padEnd(m+3),m=Math.max(2,...l=a.map(s=>s.length))&~1)

Try it online!

Groovy, 187, 176, 156, 150 bytes

f={n->m=n*.size().max()|1;h=' '*(m/2);'30734715746756276647665'*.toLong().sum{(n*.center(m+2)+[' '*(m+2),"@${'-'*m}@","|$h@$h|","|$h|$h|",'n'])[it]}}

Try it online!

(note: the tio groovy interpreter could not handle indexing Lists using Long values even though groovy 2.5.6 can. Thus the tio answer is using *.toShort() instead of *.toLong() which adds a byte)

Defines a closure f which can be called via:

println(f(['tom','ann','sue']))

where f returns a string.

Explanation:

Unobfuscating the code, we have:

f={n->

  m=n*.size().max()|1

  h=' '*(m/2)

  a=n*.center(m+2)+[' '*(m+2),"@${'-'*m}@","|$h@$h|","|$h|$h|",'n']

  '30734715746756276647665'*.toLong().sum{a[it]}

}

• 
  f={n-> - define closure f with one in-param n
  


• 
  m=n*.size().max()|1 - find max name len, binary-or to odd number


• 
  h=' '*(m/2) - h will contain floor(m/2) spaces, used later


• 
  a=...- creates an encoding list with elements:
  
  
  
  
  • indexes 0,1,2 - names, centered to max len
  
  
  • index 3 - m+2 spaces
  
  
  • index 4 - @---@ pattern, padded to len
  
  
  • index 5 - | @ | pattern, padded to len
  
  
  • index 6 - | | | pattern, padded to len
  
  
  • index 7 - newline
  
  
  
  


• 
  '307...'*.toLong().sum{a[it]}- use indicies into the encoding list to build the result. .sum uses the fact that string + string in groovy is valid.


• note that the expression '3073...'*.toLong() uses the *. spread operator to call toLong() on each character, returning a list of numbers.


• note in the answer the variable a has been inlined,nelines removed etc.

Canvas, 45 bytes

ｒ３５１⁰｛|＊@；∔；Ｊ└ｌ２Ｍ２％±├ ××ｌ⇵╷-×└＋-α∔ｋ＋│∔⇵；｝┐＋＋⇵

Try it here!

Explanation:

r    Center the input, preferring left. Converts to an ASCII-art object

     which pads everything with spaces. This is the bulk of the magic.



 251⁰{ .... }            for each number in [2, 5, 1]:

      |*                 repeat "|" vertically that many times

        @;∔              prepend an "@" - a vertical bar for later

           ;             swap top 2 stack items - put the centered art on top

            J            push the 1st line of it (removing it from the art)

             └           order the stack to [remaining, "@¶|¶|..", currentLine]

              l          get the length of the current line

               2M        max of that and 2

                 2%      that % 2

                   ±├    (-that) + 2

                      ×× prepend (-max(len,2)%2) + 2 spaces

l                 get the length of the new string

 ⇵╷               ceil(len / 2) -1

   -×             repeat "-" that many times - half of the podiums top

     └            order stack to [art, currLine, "@¶|¶|..", "----"]

      +           append the dashes to the vertical bar = "@-----¶|¶|.."

       -α∔        vertically add "-" and the original vertical bar - "-¶@¶|¶|.."

          k       remove the last line of that to make up for the middles shortness

           +      and append that horizontally - half of the podium without the name

            │     palindromize the podium

             ∔    and prepend the name

              ⇵   reverse vertically so the outputs could be aligned to the bottom

               ;  and get the rest of the centered input on top

Finally, 

┐     remove the useless now-empty input

 ++   join the 3 podium parts together

   ⇵  and undo the reversing

Abuses "and it doesn't need to be consistent", making it pretty unintelligible.

Python 2, 197 190 bytes

n=input()

w=max([3]+map(len,n))

w+=~w%2;W=w+2

S=('{:^%d}'%w).format

x,y,s='@| '

a,b,c=map(S,n);A,B,C=x+'-'*w+x,y+S(x)+y,y+S(y)+y

for l in-~W*s+a,s*W+A,s+b+s+B,A+C,B+C+s+c,C+C+A,C+C+B:print l

Try it online!

_{-6 bytes, thanks to Andrew Dunai}

Python 2, 157 bytes

a=input()

i=7

while i:print''.join(([a[k/2]]+list('-@||||    '))[7-i-k].center(max(map(len,a))|1,'- '[i+k!=6]).join('@ |'[cmp(i+k,6)]*2)for k in(2,0,4));i-=1

Try it online!

Charcoal, 63 bytes

≔÷⌈ＥθＬι²ηＦ³«Ｊ×ι⁺³⊗η⊗﹪⁻¹ι³⟦◧§θι⁺⊕η⊘⊕Ｌ§θι⟧≔⁻⁷ⅉιＰ↓ι@ηＰ↓ιＰ↓@@¹ηＰ↓ι@

Try it online! Link is to verbose version of code. Explanation:

≔÷⌈ＥθＬι²η

Calculate the number of spaces in each half of a podium.

Ｆ³«

Loop over each place. Note that input is expected to be in the order 2nd, 1st, 3rd.

Ｊ×ι⁺³⊗η⊗﹪⁻¹ι³

Position to the start of the line that will have the text.

⟦◧§θι⁺⊕η⊘⊕Ｌ§θι⟧

Output the text with enough left padding to centre it.

≔⁻⁷ⅉι

Get the height of the podium.

Ｐ↓ι@ηＰ↓ιＰ↓@@¹ηＰ↓ι@

Draw the podium.

Alternative approach, also 63 bytes:

≔÷⌈ＥθＬι²ηＦ³«Ｊ×ι⁺³⊗η⊗﹪⁻¹ι³⟦◧§θι⁺⊕η⊘⊕Ｌ§θι⪫@-@×-η⟧Ｅ⁻⁷ⅉ⪫⪫||§|@¬κ× η

Try it online! Link is to verbose version of code. Explanation:

≔÷⌈ＥθＬι²η

Calculate the number of spaces in each half of a podium.

Ｆ³«

Loop over each place. Note that input is expected to be in the order 2nd, 1st, 3rd.

Ｊ×ι⁺³⊗η⊗﹪⁻¹ι³

Position to the start of the line that will have the text.

⟦◧§θι⁺⊕η⊘⊕Ｌ§θι

Output the text with enough left padding to centre it.

⪫@-@×-η⟧

Also output the top of the podium by inserting -s between the characters of the string @-@ to reach the correct width.

Ｅ⁻⁷ⅉ⪫⪫||§|@¬κ× η

Print the remainder of the podium by spacing the |s appropriately, except that the middle character is a @ on the first row.

R 308 302 299

-6 bytes thanks to @JAD

-3 bytes thanks to @Guiseppe, now I'm under 300

function(a){i=max(1,nchar(a)%/%2)

e=2*i+1

`~`=rep

g=' '

s=g~e+2

b='@'

p='|'

t=c(b,'-'~e,b)

x=c(p,g~i,b,g~i,p)

h=sub(b,p,x)

a=lapply(a,function(q){r=nchar(q);l=(e-r)/2+1;if(r%%2<1)c(g~l,q,g~l+1)else c(g~l,q,g~l)})

cat(s,a[[1]],s,s,t,s,a[[2]],x,s,t,h,s,x,h,a[[3]],h,h,t,h,h,x,fill=length(t)*3,sep='')}

There's probably a better way to create the layout; I should try what options I have for data frames.

Try it online

Clean, 209 bytes

import StdEnv,Data.List

k=[' @|||||']

$l#m=max(maxList(map length l))3/2*2+1

=flatlines(transpose[(spaces(n*2)++g)%(0,6)\n<-[1,0,2],g<-[k:[[c,'-':tl if(i==m/2)k['  '..]]\c<-cjustify m(l!!n)&i<-[0..]]]++[k]])

Try it online!

Python 2, 188 bytes

a,b,c=l=input()

s=max(map(len,l))/2or 1

A='@'+'--'*s+'-@'

D=(' '*s).join

C=D('|@|')

D=D('|||')

d=str.center

S=s*2+3

for l in' '*S+d(a,S),' '*S+A,d(b,S)+C,A+D,C+D+d(c,S),D+D+A,D+D+C:print l

Try it online!

-5 thanks to TFeld.

Go, 436 bytes

Go is terrible for golf. But:

package main;import ("fmt";"os");func main(){;z:=os.Args;f:=3;for i:=1;i<4;i++{;if len(z[i])>f{;f=len(z[i]);};};f+=1-f%2;p:=(f-1)/2+1;b:="@";for j:=0;j<f;j++{;b+="-";};b+="@";x:=fmt.Sprintf("|%*v%*v",p,"@",p,"|");y:=fmt.Sprintf("|%*v%[1]*v",p,"|");fmt.Printf("%*v%*vn%*v%vn%*v%*vn%v%vn%v%v%*vn%v%v%vn%v%v%v",f+2,"",p+1+len(z[1])/2,z[1],f+2,"",b,p+1+len(z[2])/2,z[2],2*f+3-p-len(z[2])/2,x,b,y,x,y,p+1+len(z[3])/2,z[3],y,y,b,y,y,x)}

Broken down:

package main

import (

  "fmt"

  "os"

)

func main() {

  z:=os.Args

  f:=3

  for i:=1;i<4;i++{

    if len(z[i])>f{

      f=len(z[i])

    }

  }

  f+=1-f%2

  p:=(f-1)/2+1

  b:="@"

  for j:=0;j<f;j++{

    b+="-"

  }

  b+="@"

  x:=fmt.Sprintf("|%*v%*v",p,"@",p,"|")

  y:=fmt.Sprintf("|%*v%[1]*v",p,"|")



  fmt.Printf("%*v%*vn%*v%vn%*v%*vn%v%vn%v%v%*vn%v%v%vn%v%v%v",

  f+2,"",p+1+len(z[1])/2,z[1],

  f+2,"",b,

  p+1+len(z[2])/2,z[2],2*f+3-p-len(z[2])/2,x,

  b,y,

  x,y,p+1+len(z[3])/2,z[3],

  y,y,b,y,y,x)

}

Java 8, 399 394 373 Bytes

This solution is probably way too long, but it is a solution :)

static String r(Stringp){String s="";int l=new int{p[0].length(),p[1].length(),p[2].length()},m=Math.max(l[0],Math.max(l[1],l[2]))+2,i,q,j,a,k,t;if(m%2==0)m++;if(m==3)m=5;for(i=0;i<7;i++){for(q=0;q<3;q++)for(j=0;j<m;j++){a=(2*q+1)%3;k=2*a;t=(m-l[a])/2;s+=i==k?j>=t&j<t+l[a]?p[a].charAt(j-t):" ":i==k+1?j%(m-1)==0?"@":"-":i>=k+2?j%(m-1)==0?"|":j==m/2?i==k+2?"@":"|":" ":" ";}s+="n";}return s;}

Saved 5 Bytes by directly iterating in order (a=1,0,2 instead of q=0,1,2; a=f(q))

static String r(Stringp){String s="";int l=new int{p[0].length(),p[1].length(),p[2].length()},m=Math.max(l[0],Math.max(l[1],l[2]))+2,i,j,a,k,t;if(m%2==0)m++;if(m==3)m=5;for(i=0;i<7;i++){for(a=1;a<4;a=a==1?0:a+2)for(j=0;j<m;j++){k=2*a;t=(m-l[a])/2;s+=i==k?j>=t&j<t+l[a]?p[a].charAt(j-t):" ":i==k+1?j%(m-1)==0?"@":"-":i>=k+2?j%(m-1)==0?"|":j==m/2?i==k+2?"@":"|":" ":" ";}s+="n";}return s;}

Saved 21 Bytes thanks to @KevinCruijssen:

static String r(Stringp){String s="";int l=new int[3],m=0,i,j,a,k,t;for(String x:p)l[m++]=x.length();m=Math.max(l[0],Math.max(l[1],l[2]))+2;m+=m%2<1?1:m==3?2:0;for(i=0;i<7;i++,s+="n")for(a=1;a<4;a=a==1?0:a+2)for(j=0,k=2*a,t=(m-l[a])/2;j<m;j++)s+=i==k?j>=t&j<t+l[a]?p[a].charAt(j-t):" ":i==k+1?j%~-m<1?"@":"-":i>=k+2?j%~-m<1?"|":j==m/2?i==k+2?"@":"|":" ":" ";return s;}

As @KevinCruijssen suggested, one could also use var instead of String in Java 10+ and save some extra Bytes. I don't do this for the simple reason that I don't have Java 10 yet :D also, lambdas could be used. But this would only reduce the amount of bytes if we would leave out assigning it to a Function<String,String> variable.

In expanded form:

static String r(Stringp){

    String s=""; //The string that will be returned

    int l=new int[3], //An array containing the lengths of our three names

            m=0, //tmp variable for filling l

            i,j,a,k,t; //some declarations to save a few bytes lateron

    for(String x:p) l[m++]=x.length();

    m=Math.max(l[0],Math.max(l[1],l[2]))+2;

    m+=m%2<1? //ensure odd length of the podests

        1

        :m==3?2:0; //ensure the length is at least 3 (in my code, m is the length of the podests + 2)

    for(i=0;i<7;i++,s+="n") //iterate by row

        for(a=1;a<4;a=a==1?0:a+2) //iterate by place: a=1,0,2

            for(j=0,k=2*a,t=(m-l[a])/2;j<m;j++) //iterate by column

                //k is the row number from top in which the a-th name goes

                //t is the column at which the name starts

                //now, append the right char:

                s+=i==k? //write the name

                    j>=t&j<t+l[a]?

                        p[a].charAt(j-t)

                        :" "

                    :i==k+1? //write the top of the podest ("@---@")

                        j%~-m<1?

                            "@"

                            :"-"

                    :i>=k+2? //write the bottom of the podest ("|  |@  |")

                        j%~-m<1? //the left and right edge of the podest

                            "|"

                            :j==m/2? //the center of the podest

                                i==k+2? //are we at the first row of the bottom?

                                    "@" //the case where we have to write "| @ |"

                                    :"|" //the case "| | |"

                                :" "

                        :" "

                ;

    return s;

}

The input has to be given as String-array of length 3. An example looks like this:

public static void main(String args){

    System.out.print(r(new String{"Anthony", "Bertram", "Carlos"}));

}

Output:

          Anthony          

         @-------@         

 Bertram |   @   |         

@-------@|   |   |         

|   @   ||   |   | Carlos  

|   |   ||   |   |@-------@

|   |   ||   |   ||   @   |

PHP, 147 bytes

golfed 93 bytes off my initial idea, a straight forward <?=:

for(;~$v=_616606256046543440445[++$i];)echo$b="@   || "[$v],str_pad(($v&4?"|@":$argv)[$v&3],max(array_map(strlen,$argv))," -"[!$v],2),$b,"

"[$i%3];

takes names from command line arguments. Run with -nr or try it online.

Requires PHP 7; yields warnings in PHP 7.2 (and later, presumably). See the TiO for a +5 byte fix.

mapping:

0:@---@     = top border

1,2,3       = $argv with spaces

4: "| | |"  = default

5: "| @ |"  = below top

6: "     "  = empty

breakdown:

for(;~$v=_616606256046543440445[++$i];)echo # loop through map:

    $b="@   || "[$v],                       # print left border

    str_pad(                                # print padded string:

        ($v&4?"|@":$argv)[$v&3],                # string to be padded

        max(array_map(strlen,$argv)),           # pad length = max argument length

        " -"[!$v],                              # pad with: dashes if top border, spaces else

        2                                       # option: center text (pad on both sides)

    ),

    $b,                                     # print right border

    "n"[$i%3]                              # add linebreak every three items

;

The pre-increment for $i saves me from any tricks for the newlines.

The blank for 6 can also be empty; so I did that.

But using $argv[0] for the top border string - was the nicest golf ever. (and saved 9 bytes!)

C(GCC) 302 293 292 289 287 bytes

-6 byte thanks to ceilingcat

#define P(h,n)r=w+2-L[n];i/h?printf("%*s%*s%*s",-~r/2,D,L[n],i>h?D:N[n],r/2,D):printf(~i/h?"@%s@":"|%*s%c%*s|",~i/h?D+1:w/2,D,i>h-3?64:'|',w/2,D);

f(int**N){char L[3],w=3,i=3,r,D[99]={};for(;i--;)w=w<(L[i]=strlen(N[i]))?L[i]|1:w;memset(D+1,45,w);for(i=7;i--;puts(D)){P(4,1)P(6,0)P(2,2)}}

Run it here

Ungolfed and explained (technically you can't have comments after back slashes in macros, so this won't run)

#define P(h,n)

    r=w+2-L[n]; //get leftover width

    i/h? //if i >= h

        printf("%*s%*s%*s",-~r/2,D,L[n],i>h?D:N[n],r/2,D)://if too high print all spaces, otherwise center the name

        printf(~i/h?"@%s@":"|%*s%c%*s|",~i/h?D+1:w/2,D,i>h-3?64:'|',w/2,D);

//if (i == h - 1) print top row using D calculated if row right below top, else print '@'(64) in center, otherwise '|'

f(int**N){

    char

        L[3],//lengths of each string

        w=3,//width (init to minimum)

        i=3,//index, used in for loops

        l,//left padding

        r,//right padding

        D[99]={};//string of '-' of correct length (max 99) but first char is null for empty string

    for(;i--;)//i was set to 3 before, i will be {2,1,0}

        w=w<(L[i]=strlen(N[i]))?//set length to str len and compare to longest width so far

            L[i]|1://set it to length if longer, but make sure it is odd

            w;//do not replace

    memset(D+1,45,w); //set the first w bits of D to '-', leaves a null terminator

    for(i=7;i--;puts(D)){//i will be {6...0}

        P(4,1)//print second place, 4 high

        P(6,0)//print first place, 6 high

        P(2,2)//print thrid place, 2 high

    }

}

Here is the calling code

int main()

{

  char* N[3] = {"Tom", "Anne", "Sue"} ;

  f(N);

}

and output

         Tom         

       @-----@       

  Anne |  @  |       

@-----@|  |  |       

|  @  ||  |  |  Sue  

|  |  ||  |  |@-----@

|  |  ||  |  ||  @  |