Let $(M,g,J)$ be an Almost Hermitian manifold and $\nabla $ the Levi-Civita connection. If $$(\nabla_XJ)Y=(\nabla_YJ)X$$ for any $X,Y\in \Gamma(TM)$, can we get $\nabla J=0$, i.e., $(M,g,J)$ is Kahlerian?
The condition says that $\nabla J$ is symmetric, and I don't think it follows from the symmetry of $\nabla J$ that $\nabla J=0$. However, I can't an example to illustrate it.
Yes, $(M,g)$ is Kahler. Here's a very simple argument: consider the expression $$T(X,Y,Z) = g((\nabla_XJ)Y,Z).$$
We have that $T$ is symmetric on the first pair of arguments by your assumption, and it is skew-symmetric in the second pair of arguments in general (because since $g$ is Hermitian, $J$ is skew-adjoint, so $\nabla g = 0$ implies that each covariant derivative $\nabla_XJ$ is also skew-adjoint). This implies that $T = 0$ for very fundamental algebraic reasons (a one line computation playing with permutations, really), and thus $\nabla J = 0$.
(I should probably point out that this $T$ is, not surprisingly, the covariant differential of the Kahler form $\omega = g(J\cdot,\cdot)$ and that, under $\nabla g = 0$, the conditions $\nabla J= 0$ and $\nabla \omega = 0$ are equivalent. So I'm not pulling $T$ out of a hat, it is a very natural expression to consider.)