<s>[INST] Number of polyhedron diagonals Suppose that I have a polyhedron with given number of faces, edges and vertices are given. Is there a formula that gives me the number of polyhedron diagonals, http://mathworld.wolfram.com/PolyhedronDiagonal.html ?
 [/INST] In my comment, I failed to recognize that you were interested in just the "space diagonals". As you suspect, eliminating face diagonals requires knowing the shapes of the various faces.

Suppose, for $k = 3, 4, \dots$, there are $f_k$ faces with $k$ sides. Arguing as in my comment, we observe that the vertices of each such face determine $\frac12 k(k-1)$ segments, of which $k$ are edges; there are $\frac{1}{2} k(k-1)-k = \frac12 k(k-3)$ face diagonals per $k$-face.
Therefore, for a polyhedron with $v$ vertices, $e$ edges, and $f = f_3 + f_4 + \cdots$ faces, the number of space diagonals is given by
$$d = \frac12 v(v-1) - e - \frac12 \sum_{k=3}^{\infty}f_k\,k\,(k-3)$$
We can massage this expression a bit, using Euler's formula $v-e+f=2$ and writing $f$ in terms of the $f_k$s.
$$\begin{align}
d &= \frac12 v(v-1) - (v+f-2) - \frac12 \sum_{k=3}^{\infty} f_k k(k-3) \\[4pt]
&= 2 + \frac12 v(v-1) - v - \sum_{k=3}^\infty f_k - \frac12 \sum_{k=3}^{\infty} f_k k(k-3) \\[4pt]
&= 2 + \frac12 v(v-3) - \frac12 \sum_{k=3}^{\infty} f_k (k-1)(k-2)
\end{align}$$
This gives

$$d = \frac12\left(\;4 + v(v-3) - \sum_{k=3}^{\infty} f_k (k-1)(k-2)\;\right)$$


As a sanity check, here are some test cases:
Tetrahedron: $v = 4$; $f = f_3 = 4$; $f_{\neq 3} = 0$
$$d = \frac12\left(\; 4 + 4(4-3) - 4 (3-1)(3-2) \;\right) = 0\;\checkmark$$
Cube: $v = 8$; $f = f_4 = 6$; $f_{\neq 4} = 0$
$$d = \frac12\left(\; 4 + 8(8-3) - 6 (4-1)(4-2) \;\right) = 4\;\checkmark$$
Octahedron: $v = 6$, $f = f_3 = 8$, $f_{\neq 3} = 0$
$$d = \frac12\left(\; 4 + 6(6-3) - 8 (3-1)(3-2) \;\right) = 3\;\checkmark$$
Truncated Icosahedron (Soccer Ball): $v = 60$, $f_5  =12$, $f_6 = 20$, $f_{\text{other}} = 0$
$$d = \frac12\left(\; 4 + 60(60-3) - 12(5-1)(5-2) - 20( 6-1)(6-2) \;\right) = 1440 \;\checkmark ?$$ 
</s>