I don't know that it would be as hard as enumerating all graphs. There are more than $2^{\binom{n}2}/n!$ graphs but less than $\binom{n}{3}$ possiblities for the number of triangles. Perhaps not a lot fewer though.
$T_4=\{0,1,2,4\}.$ Removing $k=0,1,2$ disjoint edges yields $\binom{n}{3}-k(n-2)$ triangles and removing $2$ non-disjoint edges $\binom{n}{3}-2(n-2)+1.$ These will be the largest four numbers in $T_n$ so $T_5=\{0,1,2,3,4,5,7,10\}.$
It starts to seem easier to have $U_n=\{x \mid 0 \lt\binom{n}{3}-x \notin T_n \}$ so $U_5=\{1,2,4\}.$ The previous comment is that $U_n$ contains all the integers from $1$ to $2n-6$ with the exception of $n-2.$
Once $n \ge 6$ there are $4$ possible cases for removing $3$ edges yielding $\binom{n}{3}-3(n-2)+j$ for $j=0,1,2,3$ Putting this together with the previous observations and the fact that $T_5 \subset T_6$ is enough to establish that $T_6=\{0,1,2,3,4,5,6,7,8,9,10,11,12,13,16,20\}$ so $U_6=\{1,2,3,5,6\}$
I wonder how well one can do with a handful of other constructions. Certainly $T_n$ contains all the numbers $\binom{n-1}{3}+\binom{m}{2}$ for $2 \le m \le n-1$ and $\binom{n-1}{3}$ can be replaced (for $n$ not too small) by any of the larger numbers in $T_{n-1}$ (details admittedly vague here)
The reference given by "unknown", if correct (which I have no reason to doubt) gives
$U_7=\{1,2,3,4,6,7,8,11 \}$ and $U_8=\{1,2,3,4,5,7,8,9,10,13,14,19\}.$ I'll let someone else check the intermediate cases. The last one (if I have not made an error) says that $U_{12}=\{1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,21,22,23,24,25,26,31,32,33,35,41,42\}$