Here is a proof that for any fixed dimension $d$, there is a computable $n_0(d)$ such that for all $n\ge n_0(d)$ we can place the numbers $1, ..., n^d$ in a $d$-dimensional cube of side length $n$ such that any two adjacent numbers are coprime.

The main idea is to use the following lemma:

Lemma 1. If $G = (V,E)$ is a graph with maximum degree $2d$, then for any partition of the vertex set $V$ into parts $V_1, ..., V_m$ such that for each $i$ we have $|V_i| \ge 4ed$, where $e$ is the base of the natural logarithm, we can pick one vertex $v_i$ from each $V_i$ such that for $i\ne j$ vertices $v_i$ and $v_j$ are not adjacent in $G$.

Proof. This is a direct consequence of the Lovász Local Lemma. First assume WLOG that each $V_i$ contains exactly $\lceil 4ed\rceil$ elements. Suppose we randomly select one vertex $v_i$ from each set $V_i$, and for each edge $e_j = (a,b) \in E$ let $A_j$ be the event that both $a,b$ were selected. The probability of $A_j$ occurring is at most $\frac{1}{\lceil 4ed\rceil^2}$, and each event $A_j$ is dependent on at most $2(2d\lceil 4ed\rceil-1)$ other events $A_k$, so in order to apply the Local Lemma we just need to check that $e\cdot \frac{1}{\lceil 4ed\rceil^2}\cdot 2(2d\lceil 4ed\rceil-1) \le 1$, which is obvious. Thus there is a positive probability that none of the events $A_j$ occur.

Next we make a few natural definitions. From here on we set $N = n^d$.

Definition. Let $p_1, ..., p_i$ be the first $i$ prime numbers. We say that a number $v$ has $i$-type $m$ if $\text{gcd}(v,\prod_{j\le i}p_j) = m$. We'll write $\#_N^i(m)$ for the number of numbers $v$ between $1$ and $N$ with $i$-type $m$.

Definition. An $i$-type coloring of the graph $G = (V,E)$ with $|V| = N$ is an assignment of a number $m(v)$ dividing $\prod_{j\le i}p_j$ to each vertex $v$, such that for any adjacent vertices $v,w$ we have $\text{gcd}(m(v),m(w)) = 1$, and for any $m \ne 1$ the number of vertices $v$ with $m(v) = m$ is at least $\#_N^i(m)$.

The idea is to inductively convert an $(i-1)$-type coloring to an $i$-type coloring for sufficiently large $i$ using Lemma 1. To that end, we need the following lemma.

Lemma 2. There is a computable index $i_0$, depending only on $d$, such that for any $i\ge i_0$ and any $N,m$, we have $\#_N^i(mp_i) \le \max(1, \frac{1}{\lceil 4ed\rceil}\#_N^{i-1}(m))$.

Proof. If $\#_N^i(mp_i) \ge 2$, then $N/m \ge 2p_i$. If $2p_i \le N/m < p_i^3$ then we use Chebyshev's weak form of the prime number theorem to see that

$\#_N^{i-1}(m) \ge \pi(N/m) - \pi(p_i) \ge \frac{1}{\log(p_i)}\frac{N}{m}\left(\frac{7}{8}\frac{\log(p_i)}{\log(N/m)}-\frac{9}{8}\frac{p_i}{N/m}\right) \ge \frac{1}{4\log(p_i)}\frac{N}{m},$

while we trivially have $\#_N^i(mp_i) \le \frac{1}{p_i}\frac{N}{m}$. Thus if $N/m < p_i^3$ and $p_i \ge 4\lceil 4ed\rceil\log(p_i)$ then we are done.

If $N/m \ge p_i^3$, then we can, to within a constant factor, estimate the sizes of $\#_N^{i-1}(m)$ and $\#_N^i(mp_i)$ using the fundamental lemma of sieve theory, and for $i_0$ sufficiently large the first will be larger than the second by more than a factor of $\lceil 4ed\rceil$. I'll leave the details to the reader...

Now the inductive step doesn't quite work because of the existence of certain "bad pairs" $(m,p_i)$.

Definition. We call the pair $(m,p_i)$ bad if $\#_N^{i-1}(m) < \lceil 4ed\rceil$ and $\#_N^i(mp_i) = 1$ and $i \ge i_0$ and $p_i < N/2$.

We will fix this problem by taking primes between $N/2$ and $N$ and letting them pretend to be multiples of $m$ but not of $mp_i$. In order for this to work, we need to show that there are not too many bad pairs $(m,p_i)$.

Bad pairs need to satisfy all of the following constraints:

• $m$ can't have more than $\lceil 4ed\rceil-2$ prime factors: for any prime factor $q$ of $m$, $mq$ has $i$-type $m$.
• no prime factor $q$ of $m$ can be below $p_i^{\frac{1}{\lceil 4ed\rceil}}$: otherwise, all of $m, mq, mq^2, ..., mq^{\lceil 4ed\rceil}$ would have $i$-type $m$ and be below $mp_i \le N$.
• the number of primes between $p_i$ and $N/m$ must be bounded above by $\lceil 4ed\rceil$: for any prime $q$ between $p_i$ and $N/m$, $mq$ has $i$-type $m$ and is below $N$.

By the first bullet point, we see that we must have $p_i \ge N^{\frac{1}{\lceil 4ed\rceil}}$, and then by the second bullet point we see that each prime factor of $m$ must be at least $N^{\frac{1}{\lceil 4ed\rceil^2}}$. By the third bullet point and the prime number theorem, $mp_i$ must be between $N(1-O(1/\log(N)))$ and $N$ (in fact we can get much better bounds here).

Putting this all together and applying the fundamental lemma of sieve theory, we see that the number of bad pairs is at most $O(N/\log(N)^2)$, where the implied constant depends only on $d$.

For the precise description of how we artificially inflate the number of multiples of $m$ for bad pairs $(m,p_i)$, we can invent a pretending function $F_N^i(m)$ for $p_i < N/2$ satisfying the following constraints:

• $F_N^{i-1}(m) = F_N^i(m) + F_N^i(mp_i)$
• $F_N^{i-1}(m) \ge \lceil 4ed\rceil F_N^i(mp_i)$ for $i\ge i_0$.
• $F_N^i(m) \ge \#_N^i(m)$ for $m \ne 1$, $F_N^i(1) \ge 0$.

You can easily check (using the fact that $m$ has at most $\lceil 4ed\rceil$ prime factors if $m$ is part of a bad pair) that such a pretending function exists as long as $\lceil 4ed\rceil^{\lceil 4ed\rceil}$ times the number of bad pairs is at most the number of primes between $N/2$ and $N$, and for $N$ sufficiently large this will be the case. Then you can define a strong $i$-type coloring to be a coloring in which the number of vertices colored with the color $m$ is at least $F_N^i(m)$, and inductively use Lemma 1 to prove strong $i$-type colorings exist.

In order to start the inductive process, we need to come up with an $i_0$-type coloring of our hypercube (note that for $N$ large, a strong $i_0$-type coloring is the same as an ordinary $i_0$-type coloring, since there will be no bad pairs $(m,p_i)$ where $m$ has a prime factor below $p_{i_0}$). This is fairly easy as long as $n$ is large compared to $2^{i_0}$. Basically, one can start with something like this (if $n$ was $11$ and $i_0$ was 3):

$\begin{array}{ccccccccccc} 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\\ 1 & 30 & 1 & 2 & 15 & 2 & 3 & 10 & 1 & 6 & 5\\ 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\\ \vdots & & & & & & & & & & \vdots\\ 1 & 30 & 1 & 2 & 15 & 2 & 3 & 10 & 1 & 6 & 5\\ 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\end{array}$

and then fix it so there aren't quite so many $30$s, etc.