Counting xyz-graphs in $\mathbb{Z}_n^3$
Consider that hamiltonian cycles of the bipartite graph are isomorphic (that is we can always permute the rows amongst themselves and the columns similarly to reach any other hamiltonian cycle).

Consider the planes given by $(x,-,-)$ and $(-,y,-)$. Permuting these planes is exactly the same as permuting the rows and columns of the $(-,-,z)$ planes. Thus we may permute in this way to get any of the hamiltonian cycles on one of the $(-,-,z)$ planes, giving us $n!(n-1)!/2$ unique xyz-graphs. Further, we may then permute the (-,-,z) plane, fixing only the one we specifically permuted before, each of which also must give a unique xyz-graph. Thus in total we have:

$H(n) \geq \frac{n!(n-1)!^2}{2}$

Which implies the bound

$C(n) \geq \frac{1}{2}\sum\limits_{i=2}^n \frac{(n)_i^3}{i^2}\\$.

EDIT: In fact, we can do better. We have only considered sub cubes here, but we may also consider xyz graphs inside ([n],[n],[m]). Since all simple cycles are hamiltonian on $K_{n,n}$, we take $m \leq n$ to avoid double counting. Then using the same construction for our first cycle and then the same permutation process above we get:

$H(n,m) \geq \frac{n!(n-1)(m-1)!}{2}$

and then 

$C(n) \geq \frac{1}{2}\left (\sum\limits_{i,j=2 \ | \ i \geq j}^n \frac{(n)_i^2(n)_j}{ij} \right )$.