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.
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 $i$.