I wish to know if there is a situation when the Jacobi identity for the covariant derivative $\nabla_{\mu}$ $$\chi_{\mu\nu\rho}\equiv [\nabla_{\mu}, [\nabla_{\nu}, \nabla_{\rho}]]+[\nabla_{\rho}, [\nabla_{\mu}, \nabla_{\nu}]]+[\nabla_{\nu}, [\nabla_{\rho}, \nabla_{\mu}]] = 0$$ can be violated. Can anyone suggest an example where this may happen?
Edit: Just for the sake of adding some context. Consider a connection with torsion. Then, one will find that
$$[\nabla_{\mu}, \nabla_{\nu}]\varphi = 2T^{\lambda}_{~~~\mu\nu}\partial_{\lambda}\varphi$$ where $T^{\lambda}_{~~\mu\nu} = \Gamma^{\lambda}_{[\mu\nu]}$. Now, we may look at $$[\nabla_{\rho},[\nabla_{\mu}, \nabla_{\nu}]] = 2\nabla_{\rho}T^{\lambda}_{~~\mu\nu}\nabla_{\lambda}+2T^{\lambda}_{~~\mu\nu}T^{\tau}_{~~\rho\lambda}\nabla_{\tau}$$ which implies that $$\chi_{\mu\nu\rho} = 2\nabla_{\rho}T^{\lambda}_{~~\mu\nu}\nabla_{\lambda}+2T^{\lambda}_{~~\mu\nu}T^{\tau}_{~~\rho\lambda}\nabla_{\tau} + \text{cyclic}(\mu, \nu, \rho) $$ which will only be satisfied if torsion is zero. It seems the Jacobi identity is violated unless one works with a specific kind of torsion. Am I correct in my thinking?
Ok, so first of all, the Jacobi identity is always true for the covariant derivative. If $\psi$ is a section of any vector bundle on which the covariant derivative acts (indices suppressed), we have $$ [\nabla_i,[\nabla_j,\nabla_k]]\psi=\nabla_i[\nabla_j,\nabla_k]\psi-[\nabla_j,\nabla_k]\nabla_i\psi \\ = \nabla_i\nabla_j\nabla_k\psi-\nabla_i\nabla_k\nabla_j\psi-\nabla_j\nabla_k\nabla_i\psi+\nabla_k\nabla_j\nabla_i\psi. $$ If we take the cyclic sum, we get $$[\nabla_i,[\nabla_j,\nabla_k]]\psi+\text{cyclic} =\color{red}{\nabla_i\nabla_j\nabla_k\psi}+\color{green}{\nabla_j\nabla_k\nabla_i\psi}+\color{purple}{\nabla_k\nabla_i\nabla_j\psi} \\ -\color{blue}{\nabla_i\nabla_k\nabla_j\psi}-\color{brown}{\nabla_j\nabla_i\nabla_k\psi}-\nabla_k\nabla_j\nabla_i\psi \\ -\color{green}{\nabla_j\nabla_k\nabla_i\psi}-\color{purple}{\nabla_k\nabla_i\nabla_j\psi}-\color{red}{\nabla_i\nabla_j\nabla_k\psi} \\ +\nabla_k\nabla_j\nabla_i\psi+\color{blue}{\nabla_i\nabla_k\nabla_j\psi}+\color{brown}{\nabla_j\nabla_i\nabla_k\psi}. $$ The identically colored pairs cancel pairwise, hence the Jacobi identity is true for the covariant derivative.
What (apparantly) stumps OP is that if the covariant derivatives are expanded explicitly, we get - combined with the Jacobi identity - a nontrivial identity. Which is btw faulty in the OP.
For a scalar field $\phi$, the Ricci identity is $$ [\nabla_i,\nabla_j]\phi=-T^k_{\ ij}\nabla_k\phi, $$ whereas for a covector field $\omega_i$, the Ricci identity is $$ [\nabla_i,\nabla_j]\omega_k=-R^l_{\ kij}\omega_l-T^l_{\ ij}\nabla_l\omega_k. $$
We thus get $$ [\nabla_i,[\nabla_j,\nabla_k]]\phi=\nabla_i([\nabla_j,\nabla_k]\phi)-[\nabla_j,\nabla_k]\nabla_i\phi=\nabla_i(-T^l_{\ jk}\nabla_l\phi)+R^l_{\ ijk}\nabla_l\phi+T^l_{\ jk}\nabla_l\nabla_i\phi \\ =(R^l_{\ ijk}-\nabla_i T^l_{\ jk}+T^p_{\ jk}T_{\ ip}^l)\nabla_l\phi, $$ hence we get the relation (the cyclic permutation is up to a constant factor same as the antisymmetrization) $$ R^l_{\ [ijk]}=\nabla_{[i}T^l_{\ jk]}-T^l_{\ [i|p|}T^p_{\ jk]}. $$
This is nothing else than the "algebraic" Bianchi identity (which is only truly algebraic if the torsion vanishes), which has been obtained as a consequence of the Jacobi identity.