How can I calculate the minimum number of vertices needed to draw $n$ non-overlapping triangles? That is, the interiors of triangles are disjoint (equiv. the triangles disjoint except possible along their boundaries).
Details
There is no restraint on collinearity of the vertices.
I don't think $^nC_3$ is correct (as openness in triangles will not be honored then)
$n+2$ is an upper bound, since you can always take a configuration of $n$ triangles, place a vertex close to an outer edge and combine it with that edge to form another triangle. So asymptotically this number should not increase more than linearily, ruling out the $^nC_3\in O(n^3)$ you mentioned.
A lower bound can be derived from the Euler characteristic $V-E+F=\chi$ with $\chi\ge1$ (for one or more topological disks), $F=n$ and $E\ge\frac{3n}2$ resulting in $V\ge1+\frac n2$. The $E\ge\frac{3n}2$ considers three edges around each of your triangles, but each edge shared between up to two triangles.
In order to approach that bound, we should aim to keep the number of connected components to one, and also keep the number of outer edges low. Which means start with a triangle (always three outer edges) and subdivide that by adding vertices to the interior. You end up with $\chi=1$, $E=\frac{3n+3}{2}$ and $V=\frac{n+5}{2}$. Obviously this won't work for even $n$. That's because when you iterativily subdivide such a configuration, you split one big triangle into three smaller triangles, incrementing $n$ by two. So you could simply omit one of the resulting triangles of a combination with $n+1$, and end up with the final result
$$V=\left\lceil\frac{n+5}2\right\rceil$$
(You could see this as a subsequence of A008619 or A004526 if you want to.)