Building Demolition

Vim, 41 38 bytes

qw:sl 500m␍q/{␍qqjk^v$r*@wdd:-␍@w@qq@q

Here, ^ is used for a literal caret; ␍ is used for CTRL-M.

Explanation

qw:sl 500m␍q sleeps half a second, while recording the half second sleep as macro w. /{␍ moves to the floor with explosives. qq begins recording macro q, which will recursively call itself.

jk moves down and up; this generates an error if you're on the last line (ground); the error terminates the recursive macro. ^v$r* replaces everything from the first non-whitespace character to the end of the line with *'s. @w waits half a second, then dd deletes the current floor. :-␍ moves up a floor without terminating the macro if you're on the top line. @w then waits another half second, and @q calls the q macro (initially empty).

q@q stops recording the macro q, then calls it, triggering the recursion.

Animation

JavaScript (ES6), 208 198 bytes

f=
(t,s=f=>setTimeout(f,500),v=t.value.split(/(\S.*\n)/),i=v.findIndex(s=>/}/.test(s)),c=_=>t.value=v.join``,d=_=>c(v.splice(--i,2),v[3]&&s(e,i?--i:++i)),e=_=>c(v[i]=v[i].replace(/./g,'*'),s(d)))=>s(e)

<textarea id=t rows=9>
   |
  |#|
  {#}
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT
</textarea><input type=button value=Go! onclick=f(t)>

Java 7, 589 477 476 bytes

import java.util.*;void x(List<String>b,int x)throws Exception{Thread.sleep(500);int i=0,l=b.size(),z=x;String w;for(;i<l;i++){System.out.println(w=b.get(i));if(w.contains("{"))x=i;}System.out.println();w=b.get(x);i=w.contains("*")?1:0;if(i>0)b.remove(x);else b.set(x,z<0?r(w,'{','}'):r(w,'|','|'));if(l>1)x(b,i>0&x>0?x-1:x);}String r(String s,char y,char z){int a,b;return s.substring(0,a=s.indexOf(y))+s.substring(a,b=s.lastIndexOf(z)+1).replaceAll(".","*")+s.substring(b);}

Ok, it's a mess/long, but this challenge has so many annoying things for Java.. Printing multi-line; Thread.sleep(500) which requires a throws Exception; replacing a substring between two delimiters with an equal amount of *; etc.
All this causes the program to be pretty big.. It can defintely be golfed some more, maybe even halved with a different approach, but at least there is an answer now. ;)

Ungolfed:

void x(List<String>b, int x) throws Exception{
  Thread.sleep(500);
  int i = 0,
      l = b.size(),
      z = x;
  String w;
  for(;i<l; i++){
    System.out.println(w=b.get(i));
    if(w.contains("{")){
      x = i;
    }
  }
  System.out.println();
  w = b.get(x);
  i = s.contains("*")
       ? 1
       : 0;
  if(i>0){
    b.remove(x);
  }
  else{
    b.set(x, z < 0
              ? r(s, '{', '}')
              : r(s, '|', '|'));
  }
  if(l>1){
    x(b, i > 0 & x > 0
          ? x-1
          : x);
  }
}

String r(String s, chary, char z){
    int a, b;
    return s.substring(0, a=s.indexOf(y)) + s.substring(a, b=s.lastIndexOf(z) + 1).replaceAll(".", "*") + s.substring(b);
}

Test code:

import java.util.*;

class M{
  void x(List<String>b,int x)throws Exception{Thread.sleep(500);int i=0,l=b.size(),z=x;String w;for(;i<l;i++){System.out.println(w=b.get(i));if(w.contains("{"))x=i;}System.out.println();w=b.get(x);i=w.contains("*")?1:0;if(i>0)b.remove(x);else b.set(x,z<0?r(w,'{','}'):r(w,'|','|'));if(l>1)x(b,i>0&x>0?x-1:x);}String r(String s,char y,char z){int a,b;return s.substring(0,a=s.indexOf(y))+s.substring(a,b=s.lastIndexOf(z)+1).replaceAll(".","*")+s.substring(b);}

  public static void main(String[] a){
    try{
        List<String> l = new ArrayList(){{
            add("   |   ");
            add("  |#|  ");
            add("  |#|  ");
            add(" {# #} ");
            add("|# # #|");
            add("|# # #|");
            add("|# # #|");
            add("TTTTTTT");
        }};
        new M().c(l, -1);
    }
    catch(Exception e){}
  }
}

Try it here. (On ideone it outputs at once and ignores the sleep..)