Let $G$ denote a semisimple linear algebraic group over $\mathbb Q$ and let $r$ be its absolute rank.
For any prime $p$ let $v\_p$ be a vertex of the Bruhat-Tits building of $G({\mathbb Q}\_p)$ and let $D(p)$ denote the number of vertices neighboring $v\_p$. Can we say that there exists a constant $C>0$ such that
$$
D(p)\le C p^r
$$ 
holds for all $p$? If so, what is the minimal value for $C$?



Global bound for number of vertices in Bruhat-Tits building
This is not a complete answer, but only a reduction to computations with finite groups of Lie type. 


To make things simpler assume that $G$ is simple so that its building is a genuine simplicial complex. Let $X$ be the Bruhat-Tits building of $G({\mathbb Q}\_p )$ and fix a vertex $v$ of $X$. Using Bruhat-Tits theory, one attaches to $v$ a connected reductive group $G\_v$ over the residue field ${\mathbb F}\_p$. For instance when $G={\rm SL}(n)$, $G\_v ={\rm SL}(n,{\mathbb F}\_p )$ for any $v$. Then it is a classical fact that the neighbourg vertices of $v$ are in $1-1$ correspondence with the maximal parabolic subgroups of $G\_v$. Your are thus reduced to a problem of combinatorics of flag varieties over finite fields.


