The quotient $E/G$ is non-Hausdorff, I'm not sure there will be a nice geometric description.
There's a standard way to get $Y$. The symmetric space for $GL(n,\mathbb{R})$ is the symmetric space $Q$ of positive definite symmetric matrices of determinant $>0$, isomorphic to $GL(n,\mathbb{R})/O(n,\mathbb{R})$. Then $GL(n, \mathbb{Z})$ acts discretely on this space, but torsion elements have fixed points. Also, the torsion elements of $GL(n,\mathbb{Z})$ map non-trivially to $GL(n,\mathbb{Z}/p)$ for some prime $p$. One may take a $K(GL(n,\mathbb{Z}/p),1)=X$, then $GL(n,\mathbb{Z}/p)$ and therefore $GL(n,\mathbb{Z})$ acts on the universal cover $\tilde{X}$. Now, take the diagonal action of $GL(n,\mathbb{Z})$ on $Q\times \tilde{X}$. This action is free and discrete. Of course, this assumes that you have a nice way to construct $X$, which must be infinite dimensional!