We can give a construction showing that all counts up to $(1+o(1))\binom n3$ can be obtained, confirming Terry's probabilistic approach. Consider graphs consisting, apart from isolated vertices, of a sequence of cliques of size 3 or more, where each clique overlaps the previous one in two edges and otherwise has new vertices. If the cliques have size $k_1,k_2,\ldots,k_m$, then the number of triangles is $t=\sum_{i=1}^m \binom{k_i}{3}$ and the number of vertices is $n\ge 2+\sum_{i=1}^m (k_i-2)$. So the question is, for which $n,t$ do these inequalities have a solution $k_1,k_2,\ldots,k_m$?
Turn it around and ask: given $t$, what is the least $n$ for which there is a solution? (Larger $n$ are obtained by adding isolated vertices.) An upper bound is given by the greedy solution $n=N(t)$: choose $k_1$ to be as large as possible with $\binom{k_1}{3}\le t$ and continue recursively in the same manner. Since $t-\binom{k_1}{3}=O(t^{2/3})$, the $k_1$-clique dominates $N(t)$. To get an explicit bound needs some care, but wouldn't be difficult. I'm pretty sure it shows that, for large $n$, there is a solution for $0\le t\le\binom n3-O(n^2)$.
ADDED: Here are some tentative values of $[n,\Delta_n]$, where the first missing number of triangles is $\binom n3-\Delta_n$. From $n=13$ onwards, I won't swear to them. I just computed all the graphs within 16 edges of a complete graph and counted their triangles. Most seriously, I only eye-balled the output for missing values and could have missed some. The larger values are wobbling around near 0.3 $n^2$.
[4, 1], [5, 4], [6, 6], [7, 11], [8, 19], [9, 23], [10, 27], [11, 31], [12, 42], [13, 47], [14, 52], [15, 66], [16, 72], [17, 78], [18, 98], [19, 105], [20, 112]