$\DeclareMathOperator\SL{SL}$For the sake of this question, let's say that a group $G$ of finite cohomological dimension $n$ has a cohomological gap if for some $0 < i < n$ there is no subgroup $H$ of $G$ with cohomological dimension $\operatorname{cd}(H) = i$.

The question is, can an arithmetic group have a cohomological gap?

I believe the answer is no for torsion-free subgroups of $\SL_n(\mathbb{Z})$, and the same reasoning should apply for torsion-free subgroups of semisimple Chevalley groups over $\mathbb{Z}$. For $\SL_n(\mathbb{Z})$, one just needs to look inside the upper triangular subgroup to find subgroups of all dimensions. My reasoning is this generalizes to Chevalley groups as follows: the Iwasawa decomposition for such a group $G$, viewed as a linear algebraic group, gives us $G(\mathbb{R}) \sim K \times A \times N$, where $K$ is a maximal compact subgroup, $A$ is a maximal torus and $N$ is the subgroup generated by the exponentials of all positive roots. Hence, the dimension of the symmetric space $X$ associated to $G(\mathbb{R})$ is $\dim X = \dim A + \dim N$. Given that this groups split over $\mathbb{Q}$, the $\mathbb{Q}$-rank of $G(\mathbb{Z})$ is equal to $\dim A$. Therefore, as a consequence of the Borel-Serre construction, the virtual cohomological dimension of $G(\mathbb{Z})$ is $\dim N$. However, $N(\mathbb{Z})$ is a nilpotent subgroup of $G(\mathbb{Z})$ of Hirsch length $\dim N$, which gives us the bounds $$\operatorname{vcd}(G(\mathbb{Z})) = \dim N \leq \operatorname{cd}(N(\mathbb{Z}))\,.$$ Since nilpotent groups have no cohomological gap, we can find subgroups of all dimensions in $G(\mathbb{Z})$ by looking inside $N(\mathbb{Z})$.

I don't know how to approach this for general arithmetic or $S$-arithmetic groups. It's quite possible there might be a gap in "relatively small" groups such as torsion-free subgroups of $\SL_3(\mathbb{Z}[\sqrt{2}])$ or cocompact lattices in dimension $\geq 4$. This question is motivated by this related question in the Mathematics StackExchange.