Assuming that you're fine with triangles whose sides are not, themselves, parts of the diagonals, that will be fine for a heptagon whose vertices are in general convex position: we just need to subtract off triangles with two points on a diagonal.
To do this as generally as possible: if we fix a vertex $P$, the number of degenerate triangles including that vertex is $2\cdot \binom{7}{5}$ (and $2 \cdot \binom{n}{5}$ in general). The reasoning:
- We get a degenerate triangle from two intersection points on a diagonal out of $P$.
- Each intersection point lies on another diagonal with two endpoints - one to the left of the first diagonal, and one to its right.
- This determines five vertices: the other endpoint of the diagonal through $P$, and both endpoints of each of the other diagonals.
- If we choose five vertices $A_1, A_2, A_3, A_4, A_5$ in that order around the circle, the diagonal through $P$ must be $PA_3$, and the two vertices must lie either on $A_1A_4$ and $A_2A_5$, or else on $A_1A_5$ and $A_2A_4$.
We multiply this by $7$ (by $n$) for all the possible ways to choose $P$.
So you want to take your original answer of $7\cdot \binom{7}{4}\cdot \left(\binom{7}{4}-1\right)$ and subtract $2 \cdot 7\cdot \binom{7}{5}$ for the final answer.
Although you asked specifically for non-regular heptagons, it's worth pointing out that for a regular heptagon, the problem is much worse. The first concern is that there might be triple intersections of diagonals: there aren't any for the regular heptagon, but there are some for the regular $n$-gon with larger $n$. The other problem is degenerate triangles like $\triangle ABC$ in the diagram below:
It's not even immediately obvious that $A$, $B$, and $C$ are collinear (they are), let alone how to count all such triples without actually going through the diagram and dealing with every degenerate triangle one at a time.
