An example for almost commuting matrices

I'm looking for examples of the following kind:

2 sequences of unitary matrices ($$A_n,B_n$$) (where these are $$n\times n$$ matrices) that don't commute and satisfy $$||A_nB_n-B_nA_n||_1\longrightarrow_{n\rightarrow\infty}0$$ Where the $$||\cdot||_1$$ norm is the sum of the absolute values of entries.

I know that for other norms (actually for any $$p>1$$, there exists an example through a permutation matrix and a sign matrix - for example the right shift and the ordered roots of unity. This unfortunately doesn't work for the absolute value norm so I'm stuck)

I'd appreciate any help, thank you.

You could take rotation matrices in $$\mathbb{R}^3$$ which usually do not commute. If you define the matrices in such way that the angle of rotation becomes smaller when $$n$$ gets larger, the matrices will approach the identity matrix for increasing $$n$$ and the norm of the commutator will approach $$0$$, independently of the choice of the particular norm. Example: $$A_n = \begin{pmatrix} \cos\frac{\pi}{n} & -\sin\frac{\pi}{n} & 0 \\ \sin\frac{\pi}{n} & \cos\frac{\pi}{n} & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ and $$B_n = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\frac{\pi}{n} & -\sin\frac{\pi}{n} \\ 0 & \sin\frac{\pi}{n} & \cos\frac{\pi}{n} \\ \end{pmatrix}$$
Take $$A_n=\begin{bmatrix}\cos\tfrac1n&\sin\tfrac1n\\ \sin\tfrac1n&-\cos\tfrac1n\\ &&1\\ &&&\ddots&\\ &&&&1 \end{bmatrix}, \ \ \ \ \ B_n=\begin{bmatrix}-1&\\ &1\\ &&1\\ &&&\ddots&\\ &&&&1 \end{bmatrix}.$$ Then $$\|A_nB_n-B_nA_n\|_1=\left\|\begin{bmatrix}0&2\sin\tfrac1n\\-2\sin\tfrac1n&0\end{bmatrix}\right\|_1=4\,\left|\sin\tfrac1n\right|\leq\tfrac4n$$