Let $n \geq 0$ be an integer, let $A = (a_{ij})$ be the $(2n+1) \times (2n+1)$ matrix defined by $a_{ij} = 0$ unless $i + j = 2n+2$, in which case $a_{ij} = 1$. Let $G$ be the group scheme over $\mathbb{Z}$ consisting of $(2n+1) \times (2n+1)$ matrices of determinant equal to $1$ that are orthogonal for the bilinear form defined by $A$. (So $G$ is a split special orthogonal group scheme.)
Question: How does one compute the volume of $G(\mathbb{Z}_2)$ (with respect to the Haar measure)?
What I know: Let $p$ be an odd prime. Then $G$ is smooth over $\mathbb{Z}_p$, and so the map $\pi_p \colon G(\mathbb{Z}_p) \to G(\mathbb{Z}/p\mathbb{Z})$ is surjective. Let $G'_p = \ker \pi_p$. Then we have that $G(\mathbb{Z}_p) = \bigcup_{g \in G(\mathbb{Z}/p\mathbb{Z})} g \cdot G'_p$, so if we normalize the Haar measure so that $\operatorname{Vol}(G'_p) = p^{-\dim G}$, we find that $$\operatorname{Vol}(G(\mathbb{Z}_p)) = \frac{\#G(\mathbb{Z}/p\mathbb{Z})}{p^{\dim G}}.$$ Now suppose $p = 2$. The above argument fails because $G$ is not smooth over $\mathbb{Z}_2$. For example, for each $m \geq 1$, the involution $$\left[\begin{array}{ccc} 1 & 0 & 2^{m-1} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \in G(\mathbb{Z}/2^m\mathbb{Z})$$ does not admit a lift to $G(\mathbb{Z}/2^{m+1}\mathbb{Z})$. I'm not at all familiar with it, but if I understand correctly, the theory of Bruhat and Tits gives a model $\mathcal{G}$ that is smooth over $\mathbb{Z}_2$ and such that for any etale $\mathbb{Z}_2$-algebra $R$, we have that $\mathcal{G}(R) = G(R)$. In particular, this means that $\mathcal{G}(\mathbb{Z}_2) = G(\mathbb{Z}_2)$, and the image of the mod-$2$ reduction map $\pi_2\colon G(\mathbb{Z}_2) \to G(\mathbb{Z}/2\mathbb{Z})$ is equal to $\mathcal{G}(\mathbb{Z}/2\mathbb{Z})$. Am I right to then conclude that $$\operatorname{Vol}(G(\mathbb{Z}_2)) = \operatorname{Vol}(\mathcal{G}(\mathbb{Z}_2)) = \frac{\#\mathcal{G}(\mathbb{Z}/2\mathbb{Z})}{2^{\dim G}} = \frac{\#\operatorname{image}(\pi_2)}{2^{\dim G}}?$$ Is there a reference that computes $\#\operatorname{image}(\pi_2)$? According to the post https://mathoverflow.net/a/19327/76440, in the context of $2n \times 2n$ orthogonal matrices, the subgroup of $\operatorname{O}_{2n}(\mathbb{Z}/2\mathbb{Z})$ generated by the elementary orthogonal matrices "presumably" has index $2$. Since the image of $\operatorname{O}_{2n}(\mathbb{Z}_2) \to \operatorname{O}_{2n}(\mathbb{Z}/2\mathbb{Z})$ contains the elementary orthogonal matrices, I think this image also has index at most $2$.
