Clearly it's $n$ times the number of obtuse triangles with the obtuse vertex fixed at a specific point.
The angles of a regular $n$-gon have size $\frac{(n-2)\pi}n$. There are $n{-}3$ lines dividing this angle into $n{-}2$ equal parts of size $\frac\pi n$ (see the inscribed angle theorem). This means a triangle is obtuse if its remaining vertices are more than $\frac n2$ segments apart, as $\frac n2\frac\pi n=\frac\pi2$.
Therefore setting $k=\\!\\!\left\lfloor\frac n2\right\rfloor\\!{+}1$, the answer is $$n\cdot((n{-}k{-}1)+(n{-k}-2)+\ldots+1)=\frac{n(n-k)(n-k-1)}2=\frac{n\left(\left\lceil\frac n2\right\rceil\\!{-}1\right)\\!\left(\left\lceil\frac n2\right\rceil\\!{-}2\right)}2.$$