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 $\mathbf{R}^n\rtimes\mathrm{SL}_n(\mathbf{Z})$.
(Edit; added to reflect the comments:) On the other hand, $\mathbf{R}^3\rtimes\mathrm{SL}_2(\mathbf{R})$ admits cocompact lattices. Indeed, let $q=X^2+Y^2-7Z^2$. Let $G(q)(K)=K^3\rtimes\mathrm{SO}(q)(K)$. Then $q$ is $\mathbf{Q}$-anisotropic; so $G(\mathbf{Z})$ is a cocompact lattice in $G(\mathbf{R})=\mathbf{R}^3\rtimes \mathrm{SO}(q)(\mathbf{R})$. Pulling back, we get a cocompact lattice in $\mathbf{R}^3\rtimes \mathrm{SL}_2(\mathbf{R})$.