I think that the euler characteristic is 0 for the following reasons.
Firstly, the space $SL_N(\mathbb{R})$ is a bundle over the symmetric space $SO(N,\mathbb{R})\backslash SL_N(\mathbb{R}) = SP(n,\mathbb{R})=X$, the space of symmetric positive-definite real matrices of determinant 1. For a discussion of this symmetric space, see e.g. Bridson-Haefliger II.10. Then $SL_N(\mathbb{R})/SL_N(\mathbb{R})$ is a bundle over $X/SL_N(\mathbb{R})$ with fiber $SO(N,\mathbb{R})$. Note that this is an orbifold bundle, but that by passing to a torsion-free subgroup, one can assume that it is a manifold (and since you're interested in euler characteristic, this just multiplies by the index).
Now the space $X/SL_N(\mathbb{Z})$ admits a bordification by Borel-Serre. Hence $SL_N(\mathbb{R})/SL_N(\mathbb{Z})$ has a bordification by an $SO(N,\mathbb{R})$-bundle over the Borel-Serre bordification. Hence it is the interior of a manifold with boundary $M$. In this case, $H^*_c(SL_N(\mathbb{R})/SL_N(\mathbb{Z}))\cong H^*(M,\partial M)$. Then by Lefschetz duality, $\chi(H^*_c(M,\partial M))=\chi(M)$.
But since $M$ is a bundle with fiber $SO(N,\mathbb{R})$, and $\chi(SO(N,\mathbb{R}))=0$ (any Lie group has a nowhere vanishing vector field), we have $\chi(M)=\chi(SO(N,\mathbb{R}))\times \chi(X/SL_N(\mathbb{Z})) =0$, since the euler characteristic of bundles is the product of the euler characteristic of the base and the fiber.