The Held-Karp algorithm for TSP can be modified to count Hamiltonian cycles. The idea is to fix a starting vertex $v \in V$ and inductively count, for every vertex $w \neq v$ and subset $S \subseteq V$ with $v, w \in S$, the number $N(v,w,S)$ of permutations of $S$ which induce a path from $v$ to $w$.
In particular, induct on $|S|$ with a base case of $|S| = 2$, where the number of paths is 1 if $S$ is an edge and 0 otherwise (where $\Gamma(v)$ denotes the neighbourhood of $v$ and the bracket is the Iverson bracket notation):
$$ N(v,w,\{v,w\}) = [w \in \Gamma(v)] $$
and then use the same recurrence as in the Held-Karp algorithm, but taking a sum rather than a minimum:
$$ N(v,w,S) = \sum_{x \in S \setminus \{v, w \}} N(v,x,S \setminus \{w\}) [x \in \Gamma(w)] $$
You can then count the total number of Hamiltonian cycles as the number of Hamiltonian paths that begin at $v$ and end at a vertex $w \in \Gamma(v)$ (as then you can close up that path to form a Hamiltonian cycle):
$$ H = \frac{1}{2} \sum_{w \in \Gamma(v)} N(v,w,V) $$
The leading factor of $\frac{1}{2}$ is to account for the fact that we have counted directed cycles starting at $v$, and each undirected Hamiltonian cycle corresponds to two such directed cycles.
The number of subsets $S \subseteq V$ containing the starting vertex $v$ is $2^{n-1}$, and for each such subset there is an average of $\frac{n-1}{2}$ other vertices $w$, so the total number of values $N(v,w,S)$ that we need to evaluate is $(n-1) 2^{n-2}$.
For $n = 24$ (the graph in which you were interested), that is a total of 96468992 evaluations, which can easily be done on a computer.
