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 $U_6=\{1,2,3,5,6\}$


The reference given by arun, 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 $U_{12}=\{1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,21,22,23,24,25,26,31,32,33,35,41,42\}$
f

**LATER** As Gerhard notes, so far I used an enumeration of all graphs with 3 edges along with a reported enumeration of all graphs with up to $12$ nodes (although only the statistics which gave non-zero counts, not the counts themselves). However less detailed constructions might yield most of the values. Here is one:

Consider splitting $n$ nodes into two groups, of sizes $3$ and $m=n-3.$  Decree a complete graph $K_m$ on the $m$ nodes and no edges joining two of the other $3.$ Then with the between group edges one can obtain any triangle count $\binom{m}{3}+\sum_{i=1}^3\binom{m_i}{2}$ for any choices of  $m_i \le m.$ This (together with the fact that $T_{n-1} \subset T_n$) shows that $T_n$ contains all integers up to $\binom{n-2}{3}$ (and further but not all the way to $\binom{n-1}{3})$ (Any integer is a sum of three or less triangular numbers). So this gets us to $O(\binom{n}{3})$ without any enumeration. Other constructions might get further.