I was going to guess something like what David found. But a reference certainly trumps a guess. Here is some heuristic reasoning. A connected graph of course has one spanning tree. You asked about using $V$ or $E.$ Let $m(V,E)$ be the number you want and $m_2(V,E)$ the same thing with a requirement to be 2-connected.. If you only know one one them then you can't say anything more. Any time you have a bridge you can contract it to get a graph with one less edge and one less vertex. This suggests looking at $m(V,V-1+k).$ Clearly $m(V,V-1)=1.$ Using a 3-cycle with a tail, $m(V,V)=3$ while $m_2(V,V)=V.$ using $K_4$ with a tail gives 6 spanning trees for $m(V,V+1)$ which beats $9$ for two linked triangles. For $m_2(V,V+1)$ we are facing a $V$-cycle with a cord. This gives $V+s(V-s)$ spanning trees where the smaller cycle cut off by the cord has $s+1 \ge 3$ edges. So best is to take $s=2$, a triangle along with a path joining two vertices. This certainly suggests concentrating as many edges as possible on a few points.
It might be possible to describe various moves which preserve or reduce the number of spanning trees and thus characterize the minimal graphs.