I found perhaps a hint of what the relation is on page-316 of Penrose's Road to reality. For $ p- $ form $\alpha$ with index epxression, $\alpha_{b...d}$ , and a torsion free connection $\nabla_a$
$$(d \alpha)_{ab..d} = \text{anti symm}(\nabla_{a} \alpha_{b...d })$$
I compute the derivative of a two form using the above exp:
$$ \alpha = \alpha_{23} dx^2 \wedge dx^3+ \alpha_{31} dx^3 \wedge dx^1 + \alpha_{12} dx^1 \wedge dx^2$$
We must now write this in the tensor notation, by using the definition of wedge product in terms of tensor products. We have:
$$ \alpha = \alpha_{23} dx^2 \otimes dx^3 - \alpha_{23} dx^3 \otimes dx^2 + ... \text{stuff}$$
By the above, we can identify $\alpha$ in component form as $\alpha_{ij}$ with $\alpha_{ij} = - \alpha_{ji}$, we now Penrose's definition:
$$ (d \alpha)_{aij} = \text{anti symm} ( \nabla_a \alpha_{ij})= \frac13 \left[ \nabla_a \alpha_{ij} + \nabla_j \alpha_{ai} - \nabla_{i} \alpha_{aj} \right]=$$
$$ \frac13\left[ \nabla_a \alpha_{ij} + \nabla_j \alpha_{ai} - \nabla_{i} \alpha_{aj} \right] = \frac13 \left[ \frac{\partial \alpha_{ij} }{\partial x^a} -\Gamma_{ai}^p \alpha_{pj} - \Gamma_{aj}^p \alpha_{pi} + \frac{\partial \alpha_{ai} }{\partial x^j} -\Gamma_{ja}^p \alpha_{pi} - \Gamma_{ji}^p \alpha_{ap}- \frac{\partial \alpha_{aj}}{\partial x^i} + \Gamma_{ia}^p \alpha_{pj} +\Gamma_{ij}^p \alpha_{ap} \right]$$
Now due to the very convenient choice of our connection (torsion free), all the Christoffels are symmetric in lower indices and all cancel out:
$$ \left[ \nabla_a \alpha_{ij} + \nabla_j \alpha_{ai} - \nabla_i \alpha_{aj} \right] = \frac{\partial \alpha_{ij} }{\partial x^a} + \frac{\partial \alpha_{ai}}{\partial x^j } - \frac{\partial \alpha_{aj} }{\partial x^i }= \frac{\partial \alpha_{ij} }{\partial x^a} + \frac{\partial \alpha_{ai}}{\partial x^j } +\frac{\partial \alpha_{ja} }{\partial x^i }$$
Now since the differential form is a totally anti symmetric tensor evaluating at only one of the permutation determine value at all other permutation. In case of repeating indices, it is is trivially zero. So, let us evaluate at 1,2,3:
$$ (d \alpha)_{123}= \frac13 \left[ \frac{\partial \alpha_{23}}{\partial x_1} + \frac{\partial \alpha_{31}}{\partial x_2} + \frac{ \partial \alpha_{12}}{\partial x_3}\right]$$
This entry can be used to directly write down the expression in terms of the wedge product:
$$ d \alpha = \frac13 (\frac{\partial \alpha_{23}}{\partial x_1} + \frac{\partial \alpha_{31}}{\partial x_2} + \frac{ \partial \alpha_{12}}{\partial x_3}) dx_1 \wedge dx_2 \wedge dx_3$$
Seems so that I am off by a factor of $\frac13$.. where is the mistake? I've been trying to figure out what went wrong for about three days with no progress :(.
Btw for the antisymmetric part, I use the expression in comments by Deane in this post and combined the last three terms by first three using the pre-existing antisymmetric of lost two slots of the differential form.