Let a $n \times n \times n$ chessboard be given. I have just proven (by brute force) that, the queen's graph diameter, ${d_{n}^k}(Q)$ (i.e., the number of moves needed to move a queen from any 3D cell of the given chessboard to another cell belonging to the same chessboard, as the worst-case scenario), is equal to zero iff $n = 1$, ${d_{n}^3}(Q) = 1$ iff $n = 2$, ${d_{n}^3}(Q) = 2$ iff $n \in \{3,4,5\}$, and ${d_{n}^k}(Q) = 3$ $\forall n \in \mathbb{N}-\{0,1,2,3,4,5\}$.
Now, by considering any $k$-dimensional $n \times n \times n$ chessboard, it is trivial to point out that ${d_{n}^k}(Q) \leq k$ for any $n, k \in \mathbb{N}-\{0,1\}$ (since the rook's graph diameter is equal to $k$ for any $k$-dimensional chessboard such that $n \geq 2$).
My first question is as follows: "For any given $k \geq 2$, is it possible to prove the existence of a minimum value of $n \in \mathbb{Z}^+$ such that ${d_{n}^k}(Q) = k$ (e.g., if $k = 2$, then $n = 3$; if $k = 3$, then $n = 6$, and so forth)?"
Second question: "Let $n \geq 4$ and $k \geq 2$ be given. Which is the knight's graph diameter, ${d_{n}^k}(N)$, for a generic, $k$-dimensional, $n \times n \times \dots \times n$ board (e.g., ${d_{n}^k}(K)=n-1$ would describe the king's graph diameter and ${d_{n}^k}(R)=k$ returns the rook's graph diameter for the same $k$-dimensional board)?".
As a footnote, I would also note that the pawn's graph diameter is equal to $n - 2 + {d_{n}^k}(Q)$ for any $n \geq 4$, while I strongly conjecture that the knight's graph diameter, for any given $n \times n \times n$ 3D chessboard such that $n \geq 4$, is equal to $n$ (moreover, we can easily show that $\lceil{\frac{(n - 1) \cdot \sqrt{k}}{\sqrt{5}}}\rceil \leq {d_{n}^k}(N) \leq (n + 1) \cdot k$ holds for any $n$ as above and that $\lceil{\frac{(n - 1) \cdot \sqrt{k}}{\sqrt{5}}}\rceil \leq {d_{n}^k}(N) \leq n \cdot k$ is true for any odd value of $n : n > 4$, while the OEIS sequence A232007 entirely covers the planar case).
Not really an answer, just posting to revisit later.
We will write $(0,0,\dots,0)$ for one corner, to make some of the math cleaner.
I can't quite prove that minimal paths starting from $(0,0,\dots,0)$ will give the diameter, but I have an intuition for that, at least when $n$ is somewhat large.
If we are looking for the knight's distance $d$ from $(0,0,\dots,0)$ to $(a_1,a_2,\dots,a_k)$ we want to solve:
$$a_i=-2u_i-1v_i+w_i+2x_i$$
where $u_i, v_i,w_i,x_i$ are non-negative integers and $$\sum_i(u_i+x_i)=\sum_i(v_i+w_i)\tag1=d$$
You also need $v_i+w_i\leq \sum_{j\neq i} (u_j+x_j)$ or: $$u_i+w_i+v_i+x_i\leq d\tag2$$ If $(2)$ were false, then the dimension $i$ would be part of more than $d$ moves.
The obvious goal is to minimize the use of negative values.
Write $u=\sum u_i,$ and likewise for $v,w,x.$ $(1)$ can be written as $u+w=v+x=d.$
If $a_1 + \cdots + a_k=M$ and $M=3m,$ we can write $3m=m+2m,$ so we might try to get $u=v=0, w=x=m,$ and try to solve our equations with length $\ell=m.$
If $a_1 + \cdots + a_k=3m+1,$ we can write $3m+1=2(m+1)+m+(-1)$ so we might get a path of length $\ell =m+1$ with $u=0, v=1, w=m, x=m+1.$
In the case of $3m+2,$ you get $3m+2=2(m+1)+(m+2)-2,$ and we want $u=1, v=0, w=m+2, x=m+1,$ and we might find a path of length $\ell=m+2.$
Now, if $$\sum\left\lfloor \frac{a_i}2\right\rfloor\geq \ell\tag3$$ we can partition the $\ell$ $2$s amongst each dimension, and then dive out the $1s.$
For example, if $k=4, n=5$ and we are finding a path from $(0,0,0,0)$ to $(5,5,4,3)$ we get a sum of $17=3\cdot 5+2$ and thus a minimum length of $7.$ We satisfy $(3),$ so write:
$$a_1=2+2+1\\ a_2=2+2+1\\ a_3=2+2+(-2)+1+1\\ a_4=1+1+1 $$
Then we get a path:
$$(1,0,2,0)+\\ (0,1,2,0)+\\ (0,0,-2,1)+\\ (2,0,1,0)+\\ (2,0,1,0)+\\ (0,2,0,1)+\\ (0,2,0,1)=(5,5,4,3)$$
So we have the distance to $(5,4,3,2)$ is $7$.
The problem of $(3)$ rears its ugly head, though. The path from $(0,0,0,0)$ to $(1,1,1,1)$ is more complicated. Two moves is not enough. But it will still be a lot less than going to a point near the far corner, at least for relatively large $n.$ The minimal path in this case will be $$(1,2,0,0)+(0,0,1,2)+(0,1,0,-2)+(0,-2,0,1)=(1,1,1,1)$$
The maximum path will be to $(n-1,\dots,n-1)$ if $k(n-1)\equiv 2\pmod 3,$ and then the diameter will be $2+\left\lfloor \frac {k(n-1)}3\right\rfloor.$
When $k(n-1)\equiv 0,1\pmod 3$ then the diameter will be $1+\left\lfloor\frac{k(n-1)}3\right\rfloor,$ with endpoint $(n-1,n-1,\dots,n-1)$ in the $\equiv 1$ case, and $(n-1,n-1,\dots,n-1,n-2)$ in the $\equiv 0$ case.