Let $G = \operatorname{SL}_n$, and recall that its Tamagawa number is $\tau(G) = 1$ and is given by the product expansion $$\tau(G) = \operatorname{Vol}(G(\mathbb{Z})\backslash G(\mathbb{R})) \cdot \prod_p \operatorname{Vol}(G(\mathbb{Z}_p)),$$ where for each prime $p$ we have that $$\operatorname{Vol}(G(\mathbb{Z}_p)) = \frac{|G(\mathbb{F}_p)|}{p^{n^2 - 1}} = \frac{p}{p-1} \cdot \prod_{i = 0}^{n-1}(1 - p^{i-n}) = \prod_{i = 0}^{n-2}(1 - p^{i-n}).$$ I believe can show that $\operatorname{Vol}(G(\mathbb{Z}) \backslash G(\mathbb{R}))$ is bounded independent of $n$. Indeed, one can prove that $$\prod_{i = 0}^{n-2}(1 - p^{i-n})^{-1} \leq 1 + p^{-3/2}$$ by comparing the Maclaurin series expansions of the logarithms of both sides of the inequality, and $\prod_p (1 + p^{-3/2}) = \zeta(3/2)/\zeta(3)$ is a positive constant. I've checked that a similar argument works for the odd special orthogonal groups, for example.
Question: Is there any good reason to expect a priori that $\operatorname{Vol}(G(\mathbb{Z}) \backslash G(\mathbb{R}))$ is bounded independent of $n$? Is there a reference for this result in the literature?
What I know: In addition to the argument I gave above, I read in some lecture notes that when $G$ is semisimple, one does not require convergence factors in the product formula for the Tamagawa number for the product to converge, but this doesn't tell me about whether the convergence depends on the dimension of the group.
