Smooth as Teflon

Charcoal,  41 38 36 30 28 bytes

Thanks to @Emigna for helping to save two bytes, thanks to @ASCII-only for saving six bytes, and thanks to @Neil for saving another two bytes!

Ｎη↙η↑←×_η↖η↗{↗η×_η↓↘η}×=⁺³η0

Try it online!

Explanation:

Ｎη                 // Take the size of the pan as input.
↙η                 // Draw a line of η '\'s going down to the left.
↑←×_η              // Move one step up and print η underscores to the left.
↖η↗{↗η             // Print a line of η '\'s going up to the left.
                   // Then move one step north-east and print '{'.
                   // Then print a line of η '/'s going up to the right.
×_η↓               // Print '_' η times and move one step down.
↘η}                // Draw a line of η '\'s going down to the right, then print '}'.
×=⁺³η              // Print '=' η+3 times.
0                  // Print '0'

JavaScript (ES6), 171 bytes

f=
n=>(r=s=>s[0][0].repeat(n-1)+s)`  `+r`_
`+r` `.replace(/ /g," $'/$' $`$`$`\\\n")+`{`+r` `+r` `+r` }`+r`====0`+r` `.replace(/ /g,"\n $`\\$` $'$'$'/").replace(/ +\/$/,r`_/`)

<input type=number min=1 oninput=o.textContent=f(this.value)><pre id=o>

The whole pizza pan is very repetitious so the r function (designed as a tagged template literal) repeats the first character of its input n times. This handles the top and middle and lines of the pan. The rest is repeated by replacing a string of blanks; the $` and $' subsitutions automatically correspond to increasing and decreasing numbers of blanks thus positioning the / and \ appropriately. Finally the _s are filled in on the last line as it's subtly different from the second line in that respect.

JavaScript + HTML, 575 bytes (451 bytes only JS) 376 bytes (482 bytes only JS)

y=document,y.y=y.getElementById,a=(b,c)=>{w="";for(z=0;z<b;z++)w+=c;return w},d=_=>{n=Number(y.y("n").value);s="";u=" ";j="<br>",m="\\",o="/";for(i=-2;i<=2*n;i++)-2==i?s+=a(n+1,u)+a(n,"_")+j:i<n-1?s+=a(n-i-1,u)+o+a(2*(i+1)+n,u)+m+j:i==n-1?s+="{"+a(3*n,u)+"}"+a(n+3,"=")+"0"+j:i+1==2*n?s+=a(n,u)+m+a(n,"_")+o:i+1<2*n&&(s+=a(i-n+1,u)+m+a(5*n-2*i-2,u)+o+j);y.y("p").innerHTML=s};

<input type="number" id='n'><button onclick='d()'>Do</button><p id='p' style='font-family:monospace;'></p>

Not a complicated approach: several string concatenations using conditions for the five different parts of the pan: the uppermost, lowermost and middle lines and the upper and lower halves.

I shortened as much as I could, but it was the limit with this method.

EDIT: it wasn't - additionally golfed by @programmer5000