If your cone $C$ is generated by the vectors $(v_i)$, then its transverse sections are polyhedral. Lets, without losing much generality, assume that this polyhedron is closed and convex, that the $(v_i)$ are its vertices and that each face is a simplex. Each of the faces of the polyhedron, and hence of the cone, corresponds to an $(n-1)$-tuple of the vertex vectors, those generating the simplex. Let the vectors $(w_j)$ denote a set of oriented normal vectors to the faces (the normal vectors should be facing inside) of the cone $C$. These are the vectors that generate the dual cone $C^\vee$. Now, to check that $C^\vee \subseteq C$, you need to check that each of the $(w_j)$ belongs to $C$. With the $(-)^\vee$ being self-dual, this means that the inner products $(w_j,v_i)$ all need to be non-negative. If you unwind the definitions, you'll find that these inner products are determinants of matrices whose columns consist of the vector $v_i$ and the $(n-1)$-tuple of the vectors $(v_i)$ corresponding to the $w_j$, with some fixed column ordering and normalization. These determinants can also be interpreted as the signed volumes of the simplices spanned by these $n$-tuples of vectors.

In principle, once all the normalizations have been fixed, you could compute the absolute values of these volumes using only the inner products $(v_i,v_j)$, but not their signs. The reason is that any orientation reversing isometry of $V$ keeps these inner products the same, but negates all the signed volumes.