Minimum degree of the "tree graph"
If $G$ has $n$ vertices and $m$ edges, then for any spanning tree $T$ of $G$, each of the $m-n+1$ edges that are not in $T$ may be swapped with any of the edges on the path in $T$ between the endpoints of the non-tree edge. Assuming $G$ is not a multigraph, this gives at least $2(m-n+1)$ different swaps; that is, every $T$ has degree at least $2(m-n+1)$.

This bound is tight: if $G$ has a vertex $v$ adjacent to all others, and $T$ is the spanning tree consisting of all the edges incident to $v$, then the path in $T$ between the endpoints $T$ of every non-tree edge has length exactly two, so each non-tree edge participates in exactly two swaps and $T$ has degree exactly $2(m-n+1)$.

On the other hand if $G$ has girth (shortest cycle length) $g$, then the path in any tree $T$ between the endpoints of any non-tree edge, together with that edge, forms a cycle which must have length at least $g$, so the minimum degree in the tree graph must be at least $(g-1)(m-n+1)$. This bound is tight for some graphs such as the cycle graphs, and complete bipartite graphs, and Moore graphs, since these graphs contain spanning trees for which all non-tree edges induce cycles of length equal to the girth.

However, finding the minimum degree of the tree graph for an arbitrary given graph (equivalently, finding a spanning tree minimizing the sum of lengths of the cycles induced by non-tree edges) is NP-complete: see Deo, Prabhu, and Krishnamoorthy, "Algorithms for Generating Fundamental Cycles in a Graph", ACM TOMS 1982. So finding bounds such as these that are tight for all graphs appears unlikely.