Yes: it's $GL_2(\mathbf{Z}_\ell)$ itself (the quotient is trivial). Indeed, denote by $e_{ij}(x)$ (for $i\neq j$) the matrix with entry $(i,j)$ equal to $x$, diagonal entries equal to 1, and other entries 0, and $d_i(x)$, for $x$ invertible, the diagonal matrix with $(i,i)$ entry equal to $x$, and other diagonal entries equal to $1$; write $s_{ij}(x)=d_i(x)d_j(x^{-1})$. Then, for $\{i,j\}=\{1,2\}$, and $x\in \mathbf{Z}_\ell^\times$, we have $s_{ij}(x)e_{ij}(1)s_{ij}(x)^{-1}=e_{ij}(x^2)$. Since any element in $\mathbf{Z}_\ell$ is a sum of squares (of 2 squares for $\ell$ odd, 4 squares for $\ell=2$, if I remember correctly), it follows that the normal closure of $\mathrm{SL}_2(\mathbf{Z})$ contains all matrices $e_{ij}(y)$ for any $y\in\mathbf{Z}_\ell$. Since such matrices generate $\mathrm{SL}_2(\mathbf{Z}_\ell)$ (because it's a Euclidean ring), we deduce that the normal closure is all of $\mathrm{SL}_2(\mathbf{Z}_\ell)$. (The same simple argument work in $\mathrm{SL}_d$ for $d\ge 2$.)

It's the subgroup of these elements in $\mathrm{GL}_2(\mathbf{Z}_\ell)$ with determinant in $\pm 1$. Thus the quotient is naturally isomorphic to the quotient of $\mathbf{Z}_\ell^\times$ by $\{\pm 1\}$. 

Indeed, denote by $e_{ij}(x)$ (for $i\neq j$) the matrix with entry $(i,j)$ equal to $x$, diagonal entries equal to 1, and other entries 0, and $d_i(x)$, for $x$ invertible, the diagonal matrix with $(i,i)$ entry equal to $x$, and other diagonal entries equal to $1$; write $s_{ij}(x)=d_i(x)d_j(x^{-1})$. Then, for $\{i,j\}=\{1,2\}$, and $x\in \mathbf{Z}_\ell^\times$, we have $s_{ij}(x)e_{ij}(1)s_{ij}(x)^{-1}=e_{ij}(x^2)$. Since any element in $\mathbf{Z}_\ell$ is a sum of squares (of 2 squares for $\ell$ odd, 4 squares for $\ell=2$, if I remember correctly), it follows that the normal closure of $\mathrm{SL}_2(\mathbf{Z})$ contains all matrices $e_{ij}(y)$ for any $y\in\mathbf{Z}_\ell$. Since such matrices generate $\mathrm{SL}_2(\mathbf{Z}_\ell)$ (because it's a Euclidean ring), we deduce that the normal closure of $\mathrm{SL}_2(\mathbf{Z})$ is all of $\mathrm{SL}_2(\mathbf{Z}_\ell)$. (The same simple argument work in $\mathrm{SL}_d$ for $d\ge 2$.) Now since $\det(\mathrm{GL}_2(\mathbf{Z})=\{\pm 1\}$, we get the result.

(Thanks to grghxy for a correction, I was initially essentially thinking of $\mathrm{SL}_2$ and $\mathrm{GL}_2$ makes an issue with the determinant.)