Another answer is that $M_n$, the space of $n\times n$ complex matrices, carries an operator norm where the norm of a matrix is its norm as a linear operator from $\mathbb{C}^n$ to itself (giving $\mathbb{C}^n$ euclidean norm). For some of us, this is the most natural and useful norm on $M_n$.

With operator norm, $M_n$ is a finite dimensional Banach space, so it has a dual space, which is just $M_n$ equipped with trace norm. In infinite dimensions the trace class operators on $H$, with trace norm, form the (unique) predual of $B(H)$.

Another answer is that $M_n$, the space of $n\times n$ complex matrices, carries an operator norm where the norm of a matrix is its norm as a linear operator from $\mathbb{C}^n$ to itself (giving $\mathbb{C}^n$ euclidean norm). For some of us, this is the most natural and useful norm on $M_n$.

With operator norm, $M_n$ is a finite dimensional Banach space, so it has a dual space, which is just $M_n$ equipped with trace norm. In infinite dimensions the trace class operators on $H$, with trace norm, form the (unique) predual of $B(H)$.

Edit: I should add something about how this relates to the $\ell^1$ norm. The operator norm of a diagonal matrix is the $\ell^\infty$ norm of its entries, so operator norm can be seen as a sort of generalization of $\ell^\infty$ norm. Indeed, $M_n$ with operator norm contains an isometric copy of $\ell^\infty_n$ as the diagonal matrices. So it is natural that the dual norm should be the $\ell^1$ norm on the diagonal matrices.