Consequences of a $O^*(2^{n / \log(n \log n)} )$ algorithm for a #P-complete problem
Solving any old $\# P$-complete problem in $2^{O(n/\log n)}$ time would not necessarily have interesting consequences. In fact there are some $\# P$-complete problems, like counting the number of minimum vertex covers in a planar graph, which can already be done in $2^{O(n^{1/2})}$ time where $n$ is the number of vertices. (Recall that for planar graphs, the number of edges is linear in the number of vertices, so this is truly a subexponential algorithm.)

Added after Walter's recent comment: I should say that there are several $\# P$ problems which are believed to require $2^{\Omega(n)}$ time (such as counting the number of satisfying assignments to a 3-CNF formula), and the time complexities of these problems are closely related: a $2^{o(n)}$-time algorithm for one of them would imply $2^{o(n)}$-time algorithms for all of them. Some recent work in this direction can be found here. I am not 100% certain, but an $2^{O(n/\log n)}$-time algorithm for counting vertex covers in 3-regular graphs may have interesting consequences along these lines.