Counting Hamiltonian cycles in $n \times n$ square grid

The asymptotics of the number of Hamiltonian paths from one corner to the opposite (they exist only for $n$ odd) is $\tau^{n^2}$, where $\tau$ is not known exactly but satisfies $1.429 < \tau < 1.530$:

Bousquet-Mélou, M.; Guttmann, A. J.; Jensen, I., Self-avoiding walks crossing a square, J. Phys. A, Math. Gen. 38, No. 42, 9159-9181 (2005). ZBL1078.82009.

One can transform a path or cycle on $n \times n$ grid into a cycle or path on $(n+1) \times (n+1)$ grid, see figure. It follows that the number of cycles on $n \times n$ grid is between $\tau^{(n-1)^2}$ and $\tau^{(n+1)^2}$, and in any case between $1.429^{n^2}$ and $1.53^{n^2}$.

Wikipedia knows more about self-avoiding walks.

This is a summary of the sprawling comments above.

This is OEIS sequence http://oeis.org/A209077 (number of isomorphism classes of Hamiltonian circuits on $2n \times 2n$ grid). Ed Wynn discussed a method of computing these in https://arxiv.org/abs/1402.0545 and extended the known values from $1 \le n \le 4$ to $1 \le n \le 10$.

In particular, for the 6 by 6 grid (where $n=3$) which was asked about in the question, there are 149 of these circuits up to reflections and rotations.