I am trying to answer the question
Let $k[[X_1,\ldots,X_n]]$ be the power series ring in $n$ variables over a field $k$. What is the (Krull) dimension of $k[[X_1,\ldots,X_n]][(X_1\cdots X_n)^{-1}]$, where by the latter we mean the localization of $k[[X_1,\ldots,X_n]]$ at the multiplicative set $\{1,X_1\cdots X_n, (X_1\cdots X_n)^{2},\ldots \}$?
A natural guess to make is that the answer is $n-1$. And indeed I am able to prove this whenever
$$\operatorname{char}(k)\nmid n-2$$
in the following way:
It is well known that $\dim k[[X_1,\ldots,X_n]]=n$ and that $(X_1,\ldots,X_n)$ is the unique maximal ideal of $k[[X_1,\ldots,X_n]]$. Since
$$X_1\cdots X_n\in (X_1,\ldots,X_n)$$
we know that $\dim k[[X_1,\ldots,X_n]][(X_1\cdots X_n)^{-1}]\le n-1$. To show equality, we need to prove that $k[[X_1,\ldots,X_n]]$ contains a prime of height $n-1$ that contains none of the variables $X_1,\ldots,X_n$. Consider the prime $\mathfrak{p}=(X_1,\ldots, X_{n-1})$ of height $n-1$ in $k[[X_1,\ldots,X_n]]$.
Consider now a change of variables on $k[[X_1,\ldots,X_n]]$ given by mapping
$X_j$ to $(\sum_{i=1}^{n}X_i)-X_j$ for $1\le j\le n-1$ and $X_n$ to $(\sum_{i=1}^{n}X_i)$. This is an isomorphism, and checking the image of $\mathfrak{p}$, we obtain that
$$\mathfrak{p'}:=((\sum_{i=1}^{n}X_i)-X_1, \ldots, (\sum_{i=1}^{n}X_i)-X_{n-1})$$
is a prime ideal of height $n-1$ in $k[[X_1,\ldots,X_n]]$. It is easy to show that $X_j\notin \mathfrak{p'}$ for $1\le j\le n-1$ (assuming otherwise, we can show that $(X_1,\ldots,X_n)\subset\mathfrak{p'}$, contradicting what we said above about the height of $\mathfrak{p'}$).
However, it is not always true that $X_n\notin \mathfrak{p'}$, in fact this fails precisely when $\operatorname{char}(k)\mid n-2$, thus the proof does not apply for this case.
Best Answer
Yes, the result you conjectured is correct, and your proof can be easily modified to get there. We'll just pick a simpler automorphism that liberates us from the $\operatorname{char(k)} \mid n-2$ obstacle.
Namely consider the $k[[x_1, \ldots, x_n]]$-automorphism sending $x_i \rightarrow x_i + x_n$ for $i < n$ and fixing $x_n$.
This maps the height $n-1$ prime $(x_1, \ldots, x_{n-1})$ to the height $n-1$ prime $P = (x_1 + x_n, \ldots, x_{n-1} + x_n)$.
Observe that $x_k \in P$ implies $x_n \in P$ implies $x_i \in P$ for all $i$. The last statement being absurd, since $P$ is of height $n-1$, we know that $x_i \notin P$ for all $i$.