One can compute the (group cohomological) Euler characteristic of $SL_2(\mathbb{Z})$ via $$ \chi(SL_2(\mathbb{Z})) = \chi(\mathbb{Z}/2) \cdot \chi(PSL_2(\mathbb{Z})) = \frac{1}{2}\cdot (\frac{1}{2} + \frac{1}{3} - 1) = -\frac{1}{12} = \zeta(-1) $$ (since $ PSL_2(\mathbb{Z}) \cong \mathbb{Z}/2 * \mathbb{Z}/3 $). Alternatively it follows as $SL_2(\mathbb{Z})$ has $F_2$ as a $12$-index subgroup.

I was wondering whether the connection with $\zeta$ is coincidental.

The only (far fetched) connection I could come up with is that the (representation theoretic) zeta function of $SL_2(\mathbb{C})$ is the usual $\zeta$, as its irreducible representations have dimensions $1, 2, 3, ...$ (Here $\zeta_G(s) = \sum_{V\in \textrm{irreps(G)}} dim(V)^{-s}$).

Thanks! :-)

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}$One can compute the (group cohomological) Euler characteristic of $\SL_2(\mathbb{Z})$ via $$ \chi(\SL_2(\mathbb{Z})) = \chi(\mathbb{Z}/2) \cdot \chi(\PSL_2(\mathbb{Z})) = \frac{1}{2}\cdot \left(\frac{1}{2} + \frac{1}{3} - 1\right) = -\frac{1}{12} = \zeta(-1) $$ (since $ \PSL_2(\mathbb{Z}) \cong \mathbb{Z}/2 * \mathbb{Z}/3 $). Alternatively it follows as $\SL_2(\mathbb{Z})$ has $F_2$ as an index-$12$ subgroup.

I was wondering whether the connection with $\zeta$ is coincidental.

The only (far fetched) connection I could come up with is that the (representation theoretic) zeta function of $\SL_2(\mathbb{C})$ is the usual $\zeta$, as its irreducible representations have dimensions $1, 2, 3, \dotsc$. (Here $\zeta_G(s) = \sum_{V\in \operatorname{Irr}(G)} \dim(V)^{-s}$.)