This is not an answer, but a braindump of things that came up today (mainly focussing on the two-dimensional case, but most of it generalises to higher dimensions):

• In a random configuration approximately $\frac{6}{\pi^2} \approx 0.61\%$ of the adjacent cubes are coprime.

One can consider groups that naturally act on the set $C$ of configurations (such that they respect the property of a configuration being correct).

• Obviously the symmetry group of the square ($D_{4}$) acts on this set $C$.
• If $P$ denotes the set of primes $p$ such that $\frac{1}{2}n^{2} < p < n^{2}$, then $\mathrm{Sym}(P \cup \{1\})$ acts on $C$. (All numbers $1, \ldots, n^{2}$ are coprime to elements of $P \cup \{1\}$; hence we can permute this elements in a given solution.)
• Let $\mathrm{rad} \colon \{1, \ldots, N\} \to \mathbb{N}$ denote the radical function, taking $\prod_{i} p_{i}^{e_{i}}$ to $\prod_{i: e_{i}>0} p_{i}$. This function defines a partition on $\{1, \ldots, N\}$. Let $\mathrm{R}^N_{r} = \mathrm{rad}^{-1}(r)$ denote such a partition. Then $\mathrm{Sym}(\mathrm{R}^{n^2}_{r})$ acts on $C$. (This is because to numbers are coprime iff their radicals are coprime.)

I do not know a good way of estimating the size of $\mathrm{R}^{N}_{r}$, or more precisely:

Given $N$, what is (approximately) the size of the largest $\mathrm{R}^{N}_{r}$?

These groups show that there is essentially a unique solution for the $3 \times 3$ problem. One has:

• $P \cup \{1\} = \{1,5,7\}$
• $\mathrm{R}_{2} = \{2,4,8\}$
• $\mathrm{3}_{3} = \{3,9\}$.

By the parity condition mentioned in the comments, $6$ has to be at the middle of an edge, and by the $D_{4}$-action, it does not matter which edge we choose. Next, we place $2,4,8$ on the other middles of edges. Then, $3,9$ go in the corners not adjacent to $6$. Finally $P \cup \{1\}$ is placed in the remaining squares.

Ok, got to leave now. Hope this helps someone.

This is not an answer, but a braindump of things that came up today (mainly focussing on the two-dimensional case, but most of it generalises to higher dimensions):

• In a random configuration approximately $\frac{6}{\pi^2} \approx 61\%$ of the adjacent cubes are coprime.

One can consider groups that naturally act on the set $C$ of configurations (such that they respect the property of a configuration being correct).

• Obviously the symmetry group of the square ($D_{4}$) acts on this set $C$.
• If $P$ denotes the set of primes $p$ such that $\frac{1}{2}n^{2} < p < n^{2}$, then $\mathrm{Sym}(P \cup \{1\})$ acts on $C$. (All numbers $1, \ldots, n^{2}$ are coprime to elements of $P \cup \{1\}$; hence we can permute this elements in a given solution.)
• Let $\mathrm{rad} \colon \{1, \ldots, N\} \to \mathbb{N}$ denote the radical function, taking $\prod_{i} p_{i}^{e_{i}}$ to $\prod_{i: e_{i}>0} p_{i}$. This function defines a partition on $\{1, \ldots, N\}$. Let $\mathrm{R}^N_{r} = \mathrm{rad}^{-1}(r)$ denote such a partition. Then $\mathrm{Sym}(\mathrm{R}^{n^2}_{r})$ acts on $C$. (This is because to numbers are coprime iff their radicals are coprime.)

I do not know a good way of estimating the size of $\mathrm{R}^{N}_{r}$, or more precisely:

Given $N$, what is (approximately) the size of the largest $\mathrm{R}^{N}_{r}$?

These groups show that there is essentially a unique solution are essentially $4$ solutions (thanks @Carl!) for the $3 \times 3$ problem. One has:

• $P \cup \{1\} = \{1,5,7\}$
• $\mathrm{R}_{2} = \{2,4,8\}$
• $\mathrm{R}_{3} = \{3,9\}$.

1. By the parity condition mentioned in the comments, $6$ has to be at the middle of an edge, First consider the case that $6$ is at the middle of an edge, and by the $D_{4}$-action, it does not matter which edge we choose. Next, we place $2,4,8$ on the other middles of edges. Then, $3,9$ go in the corners not adjacent to $6$. Finally $P \cup \{1\}$ is placed in the remaining squares.

2. Next, observe that $6$ cannot be in the middle of the $3 \times 3$ square. After all, $\mathrm{R}_{2} \cup \mathrm{R}_{3}$ cannot be adjacent to $6$, but we can't fit those $5$ elements in the $4$ corners.

3. This leaves us with the case that $6$ is in a corner, say $(1,1)$. Observe that we cannot place an element of $\mathrm{R}_{2}$ in $(2,3)$ or $(3,2)$. Up to the aforementioned symmetries, this leaves us with $3$ possible configurations to place $\mathrm{R}_{2}$ in the squares $\{ (1,3), (2,2), (3,1), (3,3) \}$. In all cases $\mathrm{R}_{3}$ have to be placed in the squares $\{ (2,3), (3,2) \}$. The remaining squares are filled with $P \cup \{1\}$.

Ok, got to leave now. Hope this helps someone.