This is perhaps a partial answer. It's essential here to distinguish carefully between what happens when $n=2$ (where life is much more complicated) and when $n \geq 3$ (where the situation stabilizes). I'm not an expert on algebraic K-theory, but after Steinberg's papers and 1967-68 Yale lectures, the foundations were laid by Milnor in Introduction to Algebaic K-Theory (Annals of Math. Studies 72, Princeton Press, 1972). Note especially his computation in Section 10 of $K_2 \mathbb{Z}$. This group is just cyclic of order 2. In particular, when $n \geq 3$ the Steinberg group $\mathrm{St}_n(\mathbb{Z})$ is a double cover of $\mathrm{SL}_n(\mathbb{Z})$. The latter group is a lattice in the real Lie group of the same type, which is actually simply connected when $n \geq 3$. So the Steinberg group doesn't seem to be a lattice in an appropriate Lie group for your purpose.

On the other hand, when $n=2$ the computations are quite different. Here the real Lie group has an infinite cyclic covering group, which I guess might include the Steinberg group (?). But the Steinberg group itself has to be defined in a more elaborate way when $n=2$. In any case, the theory gives distinct types of presentations for $\mathrm{SL}_n(\mathbb{Z})$ in these two situations.

EDIT: This started to be an answer, but my recollections were inaccurate. It's essential here to distinguish carefully between what happens when $n=2$ (where life is much more complicated) and when $n \geq 3$ (where the situation stabilizes). The definition of the Steinberg group differs in these situations.