One answer, connecting he antisymmetry of the wedge product and the (Lie) commutator, is through the Wigner-Eckart theorem. Let $$ \hat T_{10}=\hat L_z\, \qquad \hat T_{1\pm 1}=\mp \frac{1}{\sqrt{2}} \left(\hat L_x\pm i \hat L_y\right)=\mp \frac{1}{\sqrt{2}}\hat L_{\pm} $$ Then the matrix elements $$ \langle jm'\vert \hat T_{1\mu}\vert jm\rangle=\langle jm;1\mu\vert jm'\rangle \sqrt{j(j+1)} $$ where $\langle jm;1\mu\vert jm'\rangle$ is a Clebsch-Gordan coefficient. Basically, the $-$ sign is required to define $\hat T_{11}$ as the correct $+1$ component of the tensor operator.

The connection with the wedge product is so that $$ [\hat T_{1k},\hat T_{1m}]\leftrightarrow \ue{k}\wedge \ue{m} $$ and indeed in some textbooks the commutator $[\hat T_{1k},\hat T_{1m}]$ is written as $\hat T_{1k}\wedge \hat T_{1m}$

One answer, connecting he antisymmetry of the wedge product and the (Lie) commutator, is through the Wigner-Eckart theorem. Let $$ \hat T_{10}=\hat L_z\, \qquad \hat T_{1\pm 1}=\mp \frac{1}{\sqrt{2}} \left(\hat L_x\pm i \hat L_y\right)=\mp \frac{1}{\sqrt{2}}\hat L_{\pm} $$ Then the matrix elements $$ \langle jm'\vert \hat T_{1\mu}\vert jm\rangle=\langle jm;1\mu\vert jm'\rangle \sqrt{j(j+1)} $$ where $\langle jm;1\mu\vert jm'\rangle$ is a Clebsch-Gordan coefficient. Basically, the $-$ sign is required to define $\hat T_{11}$ as the correct $+1$ component of the tensor operator.

The connection with the wedge product is so that $$ \newcommand{\ue}[1]{\hat{\mathbf{e}}_{#1}} [\hat T_{1k},\hat T_{1m}]\leftrightarrow \ue{k}\wedge \ue{m} $$ and indeed in some textbooks the commutator $[\hat T_{1k},\hat T_{1m}]$ is written as $\hat T_{1k}\wedge \hat T_{1m}$