That is, is it true that there does not exist a lattice in $G = {\rm SO}(n,1)$ which contains the group of integer points of $G$ as a proper subgroup (obviously then of finite index)? if such a lattice exists, what is the maximal possible index? I know that ${\rm SL}(n,\mathbb{Z})$ is a maximal arithmetic subgroup of ${\rm SL}(n,\mathbb{R})$ so the answer is yes if $n=2$, what is known for larger $n$? this is probably related to hyperbolic manifolds of minimal volume but I can't locate an appropriate reference... any suggestions will be appreciated :)

Here is a proof for even $n$, I am not quite sure about odd $n$ since $G=SO(n,1)$ is not an adjoint group and weak approximation might fail in this case (I do not know enough about this, you would need to ask experts, like Andrei Rapinchuk). First, note that the commensurator of $G({\mathbb Z})$ in $G({\mathbb R})$ is $G({\mathbb Q})$ (as it has to preserve the set of rational lines in the light cone: They are the fixed points of unipotent elements of $G({\mathbb Z})$). Claim. Let $G=SO(n,1)$, $n$ is even. Then $G({\mathbb Z})$ is a maximal subgroup of $G({\mathbb Q})$ and, hence, in $G({\mathbb R})$. Proof. It is known that $$ G({\mathbb Z}_p)= G({\mathbb Q}_p)\cap SL_{n+1}({\mathbb Z}_p)$$ ` is a maximal compact subgroup of $G({\mathbb Q}_p)$ for all $p$. By weak approximation (since $G$ is an adjoint group), $G({\mathbb Q})$ is dense in $G({\mathbb Q}_p)$, which implies that $G({\mathbb Z})$ is dense in $G({\mathbb Z}_p)$. If $G({\mathbb Z})$ is not a maximal subgroup of $G({\mathbb Q})$ and $G({\mathbb Z})<\Gamma< G({\mathbb Q})$ is a larger discrete subgroup, then $\Gamma$ is not contained in $G({\mathbb Z}_p)$ for some prime $p$. The closure of $\Gamma$ in $G({\mathbb Q}_p)$ is compact (since $\Gamma$ is a finite extension of $G({\mathbb Z})$) and strictly contains $G({\mathbb Z}_p)$, which contradicts maximality of $G({\mathbb Z}_p)$. qed 

