Source: Pages 89 and 90 of Sean Carroll's Spacetime and Geometry
Quite a confusion in two steps of this quantity:
$$ \begin{eqnarray} \sqrt{|g|}d^n x &=& \sqrt{|g|}dx^0 \land ... \land dx^{n-1} \newline &=& \frac{1}{n!}\sqrt{|g|} \widetilde{\epsilon}_{\mu_1 ... \mu_n}dx^{\mu_1} \land ... \land dx^{\mu_n} \newline &=& \frac{1}{n!} \epsilon_{\mu_1 ... \mu_n}dx^{\mu_1} \land ... \land dx^{\mu_n} \newline &=& \epsilon_{\mu_1 ... \mu_n}dx^{\mu_1} \otimes ... \otimes dx^{\mu_n}\newline &=& \epsilon\end{eqnarray}$$
where my confusion lies in the second equality where derivatives are reindexed via the inclusion of 1 over n factorial and the Levi-Civita symbol, as well as in fourth equality where the wedge products turn to tensor products via the suppression of the 1 over $n$ factorial.
Carroll explained the former with "since the wedge product and Levi-Civita symbol are completely anti-symmetric. (The factor of $1/n!$ takes care of the overcounting introduced by summing over permutations of the indices.)"
Carroll's definition of the wedge product (page 84) : for some p-form A and some q-form B, $$ (A \land B)_{\mu_1 ... \mu_{p+q}} = \frac{(p+q)!}{p!q!}A_{[\mu_1 ... \mu_p} B_{\mu_{p+1}... \mu_{p+q}]} $$
And so for 2 vectors we have,
$$ (A \land B)_{\mu \nu} = \frac{(1+1)!}{1!1!}A_{[\mu} B_{\nu]} $$
And then Carroll's definition of antisymmetrization (page 27) is,
$$ {T_{[\mu_1... \mu_n]}}_\rho{}^\sigma = \frac{1}{n!} (T_{\mu_1... \mu_n \rho{}^\sigma} + \text{alternating sums over permutations of indices } \mu_1...\mu_n)$$
So the wedge product continues as,
$$ (A \land B)_{\mu \nu} = \frac{(1+1)!}{1!1!}A_{[\mu} B_{\nu]} = 2 \frac{1}{2!}(A_\mu B_\nu - A_\nu B_\mu) = A_\mu B_\nu - A_\nu B_\mu$$