Tensors in general relativity This is a question on the nitty-gritty bits of general relativity.
Would anybody mind teaching me how to work these indices?
Definitions:
Throughout the following, repeated indices are to be summed over.
Hodge dual of a p-form $X$:
$$(*X)_{a_1...a_{n-p}}\equiv \frac{1}{p!}\epsilon_{a_1...a_{n-p}b_1...b_p}X^{b_1...b_p}$$
Exterior derivative of p-form $X$: $$(dX)_{a_1...a_{p+1}}\equiv (p+1) \nabla_{[a_1}X_{a_2...a_{p+1}]}$$
Given the relation 
$$\epsilon^{a_1...a_p c_{p+1}...c_n}\epsilon_{b_1...b_pc_{p+1}...c_n}\equiv p!(n-p)! \delta^{a_1}_{[b_1}...\delta^{a_p}_{b_p]}\,\,\,\,\,\,\,\,\,(\dagger)$$
where $\epsilon_{a_1...a_n}$ is an orientation of the manifold.
Why then is
$$(*d*X)_{a_1...a_{p-1}}=(-1)^{p(n-p)}\nabla^b X_{a_1...a_{p-1}b}$$?
How far I've got myself:
Firstly, I believe $(*d*X)$ means $*(d(*X))$?
$$(d*X)_{c_1...c_{n-p+1}}=\frac{n-p+1}{p!}\nabla_{[c_1}\epsilon_{c_2...c_{n-p+1}]b_1...b_p}X^{b_1...b_p}$$
Then $$*(d*X)_{d_1...d_{p-1}}=\frac{n-p+1}{(n-p+1)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$
$$=\frac{1}{(n-p)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$
Now I know that I should apply $(\dagger)$ but I don't know how to given the antisymmetrisation brackets. Would someone mind explaining it to me please? Thank you!
 A: 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.
A: You got to $$*(d*X)_{d_1...d_{p-1}}=\frac{n-p+1}{(n-p+1)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$
$$=\frac{1}{(n-p)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$
Lets denote the index $c_{n-p+1}$ by $e$ instead (or rename the indices otherwise...). Then
\begin{align}
\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p} 
&= \epsilon_{d_1...d_{p-1}ec_1...c_{n-p}}\nabla^{[e}\epsilon^{c_1...c_{n-p}]b_1...b_p}  \\
&= (-1)^{p(n-p)}\epsilon_{d_1...d_{p-1}ec_1...c_{n-p}}\epsilon^{b_1...b_p[c_1...c_{n-p}}\nabla^{e]}\\
&= (-1)^{p(n-p)}p!(n-p)!\delta^{b_1}_{[d_1}\dots \delta^{b_{p-1}}_{d_{p-1}}\delta^{b_p}_{e]}\nabla^{e}
\end{align}
(modulo some details, which you'll still need to fill in I guess...) It follows that \begin{align}
\ast (d\ast X)_{d_1 \dots d_{p-1}} 
&= (-1)^{p(n-p)}\delta^{b_1}_{[d_1}\dots \delta^{b_{p-1}}_{d_{p-1}}\delta^{b_p}_{e]}\nabla^{e}X_{b_1 \dots b_p} \\
&= (-1)^{p(n-p)} \nabla^e X_{d_1\dots d_{p-1}e}.
\end{align}
