So I have the following graph $G = (V, E)$ with $V = [d]^n$ and $E = \{\{(a_1,\dots , a_n), (b_1, \dots , b_n)\} | a_2 = b_1 a_3 = b_2, \dots a_n = b_{n-1}\}$. That is called a $(n, d)$-dimensional graph. For example, such a graph with $2$ symbols $0,\ 1$ and $3$ dimensions has $2^3 = 8$ vertices. Like: 
So here the maximal distance is $n = 3$. For the same graph with $n = 2$, the maximal distance is again $n = 2$. But my question is, is it in general always the distance $D$ between any two vertices $u, v \in V$ with $D(u,v) \leq n$ and respectively $D_{max} (u,v) = n$? If yes, how I can prove it?
Let $V$ be the set of words of length $n$ on the symbol set $S=\{0,1,\ldots,d-1\}$ (so $V=[d]^n$ in your notation).
You can get from any vertex $u$ to any vertex $v$ in $n$ steps.
Indeed, if $u=u_1u_2\cdots u_n$, and $v=v_1v_2\cdots v_n$, where $u_i,v_j\in S$ for all $i,j$, then such a sequence for length $n$ that brings you from $u$ to $v$ is: $$u_1u_2\cdots u_n \to u_2u_3\cdots u_nv_1\to u_3u_4\cdots u_nv_1v_2\to\cdots\to u_nv_1v_2\cdots v_{n-1}\to v_1v_2\cdots v_n$$
It is possible that there are shorter paths from $u$ to $v$, but there will be a pair of vertices for which the shortest distance between them is $n$: Take $u=00\cdots 00$ ($n$ zeroes) and take $v=11\cdots 11$ ($n$ ones). Any path from $u$ of length $k<n$ will end at a vertex, whose corresponding string has $n-k$ leading zeroes, and you can thus never reach $v$ from $u$ in less that $n$ moves.