The polyhedron described looks like this, a truncated cuboctahedron.

Truncatedcuboctahedron.jpg
The number of segments joining the vertices of the polyhedron is ${48\choose2} = 1128$. We must now subtract out those segments that lie along an edge or a face.
Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$.
Each vertex is formed by the intersection of 3 edges. Since every edge is counted twice, once at each of its endpoints, the number of edges $E$ is $\frac{3}{2}V = 72$.
Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes $\frac{n(n-3)}{2} = 2$ diagonals, each hexagon $9$, and each octagon $20$. The number of diagonals is thus $2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216$.
Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = \boxed{840}$.