I don't want to enter in the detail of the calculus, but the error is in the order of $\epsilon$ terms, it is not:
$$\epsilon^{\large A_pC_{n-p}} \epsilon_{\large B_pC_{n-p}}$$
but :
$$\epsilon^{\large A_pC_{n-p}} \epsilon_{\large C_{n-p}B_p}$$
Where indices like $A_p$ means $a_1...a_p$
So, before applying general formulae concerning Levi-Civita tensor, you have to invert, for instance, $B_p$ and $C_{n-p}$. If you move a term belonging to $B_p$ tp the left, you got a term $(-1)^{n-p}$. Because you have $p$ terms in $B_p$, you get a total term $(-1)^{p(n-p)}$
You have the same type of problem in looking at the difference between $*(*X)$ and $X$, so, if you want, begin by this calculus.
Secondly, you have to take in account the metrics (here I consider constant metrics), so your first formula for the Hodge dual is not true, you have to use $\sqrt{|g|} \epsilon_{AB}$ instead of $\epsilon_{AB}$, and $ - sgn(g) \frac{1}{\sqrt{|g|}} \epsilon^{AB}$, instead of $\epsilon^{AB}$. So, there is a difference between an euclidean metrics and a Minkowski metrics (You will get a minus sign with the Minkowski metrics). Do not forget that your raise or lower the indices with the metrics.
Finally, You mix $2$ things, ordinary exterior derivative $d$, and covariant exterior derivative $D$. I think, that, if you use the notation $d$, it concerns only standard derivatives.
[EDIT]
Note also, that all your factorial terms disappear if you don't take an totally antisymmetric expression, but just an ordered expression, for instance :
$\frac{1}{2!} \epsilon^{ij} X_{ij} = X_{12}$ ($X$ is supposed anti-symmetric)
[EDIT2]
Yes, (∗d∗X) means ∗(d(∗X))
[COMPLETE SOLUTION]
I will only give a solution in a constant euclidean metrics.
Multi-Indices $A_p$ means $a_1...a_p$, I used a mono-indice $C$ which means $c$.
So, we have :
$$(*X)_{\large B_{n-p}} = \frac{1}{p!} \epsilon^{\quad \quad\large A_{p}}_{\large B_{n-p}} \quad (X)_{\large A_{p}} \quad \quad \quad \quad(1)$$
$$(d*X)_{\large CB_{n-p}} = (n-p+1) \quad \partial_{\large [C}(*X)_{\large B_{n-p}]} \quad \quad \quad \quad(2)$$
$$*(d*X)_{\large D_{p-1}} = \frac{1}{(n-p+1)!} \epsilon^{\quad \quad\large CB_{n-p}}_{\large D_{p-1}} \quad (d*X)_{\large CB_{n-p}} \quad \quad \quad \quad(3)$$
So, using $(1), (2), and (3)$, you have :
$$*(d*X)_{\large D_{p-1}} = \frac {n-p+1}{(n-p+1)! p!} \epsilon^{\quad \quad\large CB_{n-p}}_{\large D_{p-1}} ~~\epsilon^{\quad \quad\large A_{p}}_{\large B_{n-p}} \quad \partial_{\large [C}(X)_{\large A_{p}]} \quad \quad \quad \quad(4)$$
That is :
$$*(d*X)_{\large D_{p-1}} = \frac {1}{(n-p)! p!} \epsilon_{\large D_{p-1}C'B_{n-p}} ~~\epsilon^{\large \large B_{n-p}A_{p}} g^{CC'} \quad \partial_{\large [C}(X)_{\large A_{p}]} \quad(5)$$
Putting $\large B_{n-p}$ at left, we get:
$$*(d*X)_{\large D_{p-1}} = (-1)^{p(n-p)} \frac {1}{(n-p)! p!} \epsilon_{\large B_{n-p}D_{p-1}C'} ~~\epsilon^{\large \large B_{n-p}A_{p}} g^{CC'} \quad \partial_{\large [C}(X)_{\large A_{p}]} \quad(6)$$
Using the contraction formulae of the Levi-Civita symbol, we get :
$$*(d*X)_{\large D_{p-1}} = (-1)^{p(n-p)} \delta_{d_1}^{[a_1} \delta_{d_2}^{a_2} ...\delta_{d_{p-1}}^{a_{p-1}} \delta^{a_p]}_{c'} g^{CC'} \quad \partial_{\large [C}(X)_{\large A_{p}]} \quad \quad(7)$$
Because, both terms $\delta_{\large D_{p-1}C'} ^{\large [A_{p}]}$ (that is $\delta_{d_1}^{[a_1} \delta_{d_2}^{a_2} ...\delta_{d_{p-1}}^{a_{p-1}}$) and $\partial_{\large [C}(X)_{\large A_{p}]}$ are antisymmetric in $\large A_{p}$, we could write :
$$*(d*X)_{\large D_{p-1}} = (-1)^{p(n-p)} \delta_{d_1}^{a_1} \delta_{d_2}^{a_2} ...\delta_{d_{p-1}}^{a_{p-1}} \delta^{a_p}_{c'} g^{CC'} \quad \partial_{\large [C}(X)_{\large A_{p}]} \quad \quad(8)$$
So, finally,
$$*(d*X)_{\large D_{p-1}} = (-1)^{p(n-p)} g^{CC'} \quad \partial_{\large [C}(X)_{\large D_{p-1}C']} \quad \quad(9)$$
With a (constant) Lorentzian metrics, with determinant $-1$, you will have a $sgn(g) = -1$ supplementary term, because, in this case, you have $\epsilon^{\large F_n} = sgn(g) \epsilon_{\large F_n} = - \epsilon_{\large F_n}$, and because contractions formulae for Levi-Civita symbol work only for euclidean metrics.
