Let $\gamma(G)$ denote the domination number of a graph, and $G\,\square\,H$ denote the cartesian product of two graphs. Then $K_8\,\square\, K_8$ is the rook graph, whose vertices are the squares of a chessboard, with edges between squares a rook can move between. We can similarly define the rook graph of any square chessboard as $K_n \,\square \, K_n$, and define $d$-dimensional chessboards as $(K_n)^{\square \,d}$, where generalized rooks move by sliding parallel to one of the coordinate axes.

Clearly, $\gamma(K_n\,\square\, K_n)=n$. Put another way, it is possible to place $n$ rooks on an $n\times n$ chessboard so every square contains a rook or is attacked by one, while $n-1$ rooks are insufficient. Much less easily, you can prove $\gamma(K_n\,\square\,K_n\,\square\,K_n)=\lceil n^2/2\rceil$, which also has an interpretation with rooks on a 3D board.

These nice answers for two and three dimensions gave me hope there was a nice one for 4, so my question is this:

Is anything known about $\gamma ((K_n)^{\square \,4})$? Or, how many rooks does it takes to dominate an $n\times n\times n \times n$ chessboard?

It's easy to show that $\gamma ((K_2)^{\square \,4})= 4$, and not hard to show $\gamma ((K_3)^{\square \,4})= 9$. But for a $4\times 4\times 4\times 4$ board, all I know is that the domination number is at least 23 and at most 32. Unfortunately, it is infeasible to brute force $\gamma((K_4)^{\square \,4})$.