The abstract definition of the curvature tensor is good as is, see, if $\partial_\mu$ is the $\mu$th coordinate basis vector for some local chart, we have $\nabla_{\partial_\mu}\partial_\nu=\Gamma^\sigma_{\mu\nu}\partial_\sigma$, so we have $$ R(\partial_\mu,\partial_\nu)\partial_\rho=\nabla_{\partial_\mu}\nabla_{\partial_\nu}\partial_\rho-\nabla_{\partial_\nu}\nabla_{\partial_\mu}\partial_\rho=\nabla_{\partial_\mu}(\Gamma^\sigma_{\nu\rho}\partial_\sigma)-\nabla_{\partial_\nu}(\Gamma^\sigma_{\mu\rho}\partial_\sigma)=\partial_\mu\Gamma^\sigma_{\nu\rho}\partial_\sigma+\Gamma^\sigma_{\nu\rho}\nabla_{\partial_\mu}\partial_\sigma-(\mu\leftrightarrow\nu)=\\=\partial_\mu\Gamma^\sigma_{\nu\rho}\partial_\sigma+\Gamma^\sigma_{\nu\rho}\Gamma^\lambda_{\mu\sigma}\partial_\lambda-(\mu\leftrightarrow\nu)=(\partial_\mu\Gamma^\sigma_{\nu\rho}+\Gamma^\sigma_{\mu\lambda}\Gamma^\lambda_{\nu\rho}-(\mu\leftrightarrow\nu))\partial_\sigma, $$ so the formula $$ R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z $$ gives the correct curvature formula in the case of torsion as well.
What confuses you is that in index notation, the expression $[\nabla_\mu,\nabla_\nu]$ actually corresponds to an antisymmetric expression made from second covariant derivatives, in abstract language, the torsionless curvature tensor can be expressed as $$ R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z, $$ and its THIS formula that needs to be modified for torsion. It's because $\nabla^2_{X,Y}Z$ corresponds to $X^\mu Y^\nu\nabla_\mu\nabla_\nu Z^\sigma$ in index notation. In abstract notation, $\nabla_X\nabla_Y Z\neq\nabla^2_{X,Y}Z$, because in the first term, $\nabla_X$ also acts on $Y$, while for the "second covariant derivatives", $Y$ is applied "from outside the differential operator".
In index notation, $$ \nabla_X\nabla_Y Z\Longleftrightarrow X^\mu\nabla_\mu(Y^\nu\nabla_\nu Z^\sigma) \\ \nabla^2_{X,Y}Z\Longleftrightarrow X^\mu Y^\nu\nabla_\mu\nabla_\nu Z^\sigma,$$ work out for yourself in component notation that these two are not equivalent.
The difference between the antisymmetric form of $\nabla_X\nabla_Y Z$ and $\nabla^2_{X,Y}Z$ will produce a term proportional to $\nabla_{\nabla_YX-\nabla_XY}$, which is $\nabla_{[X,Y]}$ iff $\nabla$ is torsionless, hence the need to modify $R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z$ if we have nonvanishing torsion, since the correct curvature tensor is given by $R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$.
This is a fact from Riemannian geometry, yes, but has nothing to do with the Hermitian structure. The key point is that you can rephrase the torsion-free condition by working with an orthonormal coframe $\theta_i$ and saying
$$d\theta_i = \sum \omega_{ij}\wedge\theta_j$$
with $\omega_{ij}=-\omega_{ji}$. (Ordinarily, you'd have $d\theta_i = \sum\omega_{ij}\wedge\theta_j + \tau_i$, where $\tau$ gives the torsion.) Here $\omega_{ij}$ gives the connection form.
Then you can check that $\nabla \theta_i = \sum \omega_{ij}\otimes\theta_j$, and the rest is immediate, since $d\theta_i = \sum\omega_{ij}\wedge\theta_j = \sum \omega_{ij}\otimes\theta_j - \theta_j\otimes\omega_{ij}$.
Best Answer
A smooth manifold $M$ may be equipped with many different geometric structures. For example:
It turns out that on a Riemannian manifold $(M,g)$, there is a "best" connection to choose, the Levi-Civita connection. The Levi-Civita connection is characterized by being both (1) compatible with the metric, and (2) torsion-free.
To reiterate: once a metric $g$ is chosen, there are many connections $\nabla$ we can work with, but the Levi-Civita is the best one (in a certain precise sense).
Given an affine connection $\nabla$, any whatseover, we can define its associated torsion tensor by $$T^\nabla(X,Y) = \nabla_XY - \nabla_YX - [X,Y].$$ We say that a connection is torsion-free iff $T^\nabla = 0$. Some connections (like the Levi-Civita connection) have this property, but others don't.
Finally, given an affine connection $\nabla$, any whatsoever, and a coordinate chart, we can talk about the associated Christoffel symbols, denoted $\Gamma^k_{ij}$. Again, they depend on both the choice of connection and the coordinate chart.
It is a fact (exercise) that $\Gamma^k_{ij} = \Gamma^k_{ji}$ in every coordinate chart if and only if $\nabla$ is torsion-free.