Covariant derivative versus exterior derivative

Three comments:

1. There are two different conventions in common use for the wedge product. The one Spivak uses and I use (which I call the determinant convention) is $$ \alpha \wedge \beta = \frac{(k+l)!}{k!l!} \operatorname{Alt}(\alpha\otimes\beta), $$ when $\alpha$ is a $k$-form and $\beta$ is an $l$-form. The one Kobayashi & Nomizu use (the Alt convention) is $$ \alpha \wedge \beta = \operatorname{Alt}(\alpha\otimes\beta). $$ The formula you wrote down for $d\omega$ is correct if you're using the determinant convention. But the formula $d\omega = \operatorname{Alt}(\nabla\omega)$ would only be correct using the Alt convention. With the determinant convention, the formula should be $d\omega = \pm (k+1)\operatorname{Alt}(\nabla\omega)$ when $\omega$ is a $k$-form.
2. The plus or minus sign depends on how you define the $(k+1)$-tensor $\nabla \omega$. Some authors define it to be $$\nabla\omega(\dots,Y) = \nabla_Y\omega(\dots),$$ while others define it to be $$\nabla\omega(Y,\dots) = \nabla_Y\omega(\dots).$$ Your computation shows that you're using the first convention, in which case the correct formula is $d\omega = (-1)^k (k+1)\operatorname{Alt} (\nabla\omega)$.
3. To derive your formula $\color{red}{(\ast)}$ (or rather the version with the correct constant multiple) by brute force, you can note that each of the terms of the form $\nabla_{X_i} X_j$ can be matched up with a term $\nabla_{X_j} X_i$ with the opposite sign, and because the connection is symmetric, these combine to give $[X_i,X_j]$, which then matches one of the terms in the formula for $d\omega$. But a much easier approach is to note that both $d\omega$ and $\operatorname{Alt}(\nabla\omega)$ are well-defined tensor fields, and thus their values at a point $p$ can be compared in terms of any convenient local frame. If you let the $X_i$'s be coordinate vector fields in Riemannian normal coordinates centered at $p$, then many terms go away.

This is only a partial answer.

This result is proved in Kobayashi-Nomizu, Foundations of differential geometry, vol. 1: it's Corollary 8.6 p.149 ($\operatorname{Alt}(\nabla w) = dw$ holds for any torsion-free connection $\nabla$). But it does not look like their proof is what you are looking for.

I understand the calculation that you are trying to do and I'm not sure how to fix it, but here is an observation that may help: if $\omega$ is a $k$-form, in other words an element of $\Gamma(\Lambda^k T^*M)$, then $\nabla \omega$ is an element of $\Gamma(T^*M \otimes \Lambda^k T^*M)$, so I think that $\operatorname{Alt}(\nabla w)$ is simply given by $\operatorname{Alt}(\nabla w) = \sigma(\nabla w)$, where $\sigma : \Gamma(T^*M \otimes \Lambda^k T^*M) \to \Gamma(\Lambda^{k+1} T^*M)$ is the linear map given by $\sigma(\alpha \otimes \eta) = \alpha \wedge \eta$. In this paper, they proceed to show the result, but they use local coordinates. It's probably possible to do a coordinate-free proof like you want but I haven't tried long enough.