I have been asked to consider the antisymmeterisation: \begin{equation} \nabla_{[\mu}\nabla_{\nu}\omega_{\lambda]}. \end{equation}
By expanding out this expression, I have found: \begin{equation} \nabla_{[\mu}\nabla_{\nu}\omega_{\lambda]} = -\frac{\omega^\sigma}{6}R_{\sigma[\lambda\mu\nu]}. \end{equation} I believe I would like for the left hand side to vanish to conclude the First Bianchi Identity: \begin{equation} R_{\sigma[\lambda\mu\nu]} = 0, \end{equation} although I do not know what to do from here.
Edit: I have been asked to show the details of how I obtained the second equation:
I expanded out the expression as follows: \begin{equation} \nabla_{[\mu}\nabla_{\nu}\omega_{\lambda]} = \frac{1}{3}(\nabla_\mu\nabla_{[\nu}\omega_{\lambda]} + \nabla_\nu\nabla_{[\lambda}\omega_{\mu]} + \nabla_\lambda\nabla_{[\mu}\omega_{\nu]}) \end{equation}
Expanding this further I got: \begin{equation} \nabla_{[\mu}\nabla_{\nu}\omega_{\lambda]} = \frac{1}{6}(\nabla_\mu\nabla_{\nu}\omega_{\lambda} - \nabla_\mu\nabla_{\lambda}\omega_{\nu} + \nabla_\nu\nabla_{\lambda}\omega_{\mu}- \nabla_\nu\nabla_{\mu}\omega_{\lambda} +\nabla_\lambda\nabla_{\mu}\omega_{\nu} -\nabla_\lambda\nabla_{\nu}\omega_{\mu}) \end{equation} Using the identity \begin{equation} [\nabla_\alpha,\nabla_\beta]\omega_{\delta} = -R^{\sigma}_{\delta\alpha\beta}\omega_\sigma = -R_{\sigma\delta\alpha\beta}\omega^\sigma \end{equation} I got: \begin{equation} \nabla_{[\mu}\nabla_{\nu}\omega_{\lambda]} = -\frac{1}{6}(R_{\sigma\lambda\mu\nu}\omega^\sigma + R_{\sigma\nu\lambda\mu}\omega^\sigma + R_{\sigma\mu\nu\lambda}\omega^\sigma) = -\frac{\omega^\sigma}{6}R_{\sigma[\lambda\mu\nu]} \end{equation}
The argument why $\nabla_{[\mu}\nabla_{\nu}\omega_{\lambda]}=0$ is as follows: As you have implicitly noted in your computations, you can alternate the inner derivative first. But then torsion-freeness of the Levi-Civita connection implies that $\nabla_{[\nu}\omega_{\lambda]}=(d\omega)_{\nu\lambda}$. For the same reason you get $\nabla_{[\mu}(d\omega)_{\nu\lambda]}=d(d\omega)_{\mu\nu\lambda}=0$. So the main ingredient is $d^2=0$.