No, there's no uniform (=cocompact) lattice in $\mathbf{R}^n\rtimes\mathrm{SL}_n(\mathbf{R})$ for any $n\ge 2$. Up to the action by automorphisms of $\mathrm{GL}_n(\mathbf{R})$, all lattices are contained in $\mathbf{R}^n\rtimes\mathrm{SL}_n(\mathbf{Z})$, which is not cocompact.

Indeed, let $\Gamma$ be a lattice. In a connected Lie group, the intersection of a lattice with the amenable radical is a lattice in the amenable radical. So $\Gamma\cap\mathbf{R}^n$ is a lattice in $\mathbf{R}^n$; hence modulo a global automorphism induced by some element of $\mathrm{GL}_n(\mathbf{R})$, we can assume that $\Gamma\cap\mathbf{R}^n=\mathbf{Z}^n$. The subgroup $\Gamma$ is then contained in the normalizer of $\Gamma\cap\mathbf{R}^n=\mathbf{Z}^n$, and this normalizer is equal to $\mathbf{R}^n\rtimes\mathrm{SL}_n(\mathbf{Z})$.