I think I found a solution and that $\{0\}$ is the only simple module (in the norm-closed sense).

To obtain that, you have to use the following theorem of Wiener, see Theorem 4.64 of Folland's

4.64 Theorem. (Wiener's Theorem), if $J$ is a closed ideal in $L^1(G)$ and $\nu(J) = \emptyset$, then $J=L^1(G)$.

where $G$ is a LCH Abelian group and $\nu(J) \subset \widehat{G}$ is the nullset over the dual. You also the following Theorem from the same book:

4.67 Theorem. Suppose $J$ is a closed ideal in $L^1(G)$, $f \in L^l(G)$, and $\nu(J) \subset \nu(f)$. If $\partial \, \nu(J) \cap \partial \, \nu(f)$ contains no non empty perfect set, then $f \in J$.

If $J = \{0\}$, it is simple. Otherwise, there are two cases, if $J = \ell^1(\mathbf{Z})$, then there is nothing to prove. If $J$ is smaller than $\ell^1(\mathbf{Z})$, then its null set $S$ has to be non-empty by Wiener's Theorem. It is also non-total. But then $\mathbf{T} \setminus S$ contains a small interval whose complement C has non-intersecting boundary with that of $S$. The annihilator of $C$ is a closed submodule.

Edited because the original answer solved the problem for $\ell^1(\mathbf{Z})$ submodules not for $\ell^2(\mathbf{Z})$ ones.

An $\ell^1(\mathbf{Z})$-submodule of $\ell^2(\mathbf{Z})$ is just an invariant subspace under the left regular representation $\lambda$. If $P:\ell^2(\mathbf{Z}) \to \ell^2(\mathbf{Z})$ is the projection onto a $\lambda$-invariant subspace, then it lies in the commutant of $\lambda[\mathbf{Z}]$, which is just $L^\infty(\mathbf{T})$ after conjugating with the Fourier transform and applying the Plancherel theorem. This gives that, after taking the Fourier transform, any closed submodule is of the form $\mathbf{1}_E \cdot L^2(\mathbf{T})$, for a measurable set $E$. The reciprocal is also true. Every measurable set (up to measure zero differences) gives a submodule. Thus, the only simple module is $\{0\}$.