Why are circular unit vectors often defined as $\hat{\mathbf e}_\pm = \mp (\hat{\mathbf e}_x \pm i \hat{\mathbf e}_y)/\sqrt{2}$

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 $\hat T_{1k}\wedge \hat T_{1m}$

$\newcommand{\ue}[1]{\hat{\mathbf{e}}_{#1}}$ If you define $\ue\pm$ as

$$\ue\pm= \frac{1}{\sqrt{2}}\big(\ue x \pm \mathrm{i} \ue y\big)$$

you get

$$\ue+\wedge \ue- = -\mathrm{i}\ue z.$$

If, instead, you define

$$\ue+ = -\frac{1}{\sqrt{2}}\big(\ue x + \mathrm{i} \ue y\big)$$

and

$$\ue- = \frac{1}{\sqrt{2}}\big(\ue x - \mathrm{i} \ue y\big),$$

the cross product becomes

$$\ue+\wedge \ue- = \mathrm{i}\ue z.$$

Therefore, in the former case, $(\ue+,\ue-,\mathrm{i}{\ue z})$ has the same orientation of $(\ue x,\ue y, \ue z)$, whereas in the latter the orientation is preserved by $(\ue -,\ue +, \mathrm{i}\ue z)$.

Is there any reason to prefer one orientation with respect to the other? I would say no, it's probably just a matter of tradition.