to Question 1: Yes.
To prove this, let me fix a positive integer $n$ and denote your matrix (whose
determinant $f_{n}$ is) by $A$. The notation $\left[ k\right] $ shall be
used for the set $\left\{ 1,2,\ldots,k\right\} $ whenever $k$ is an integer.
The notation $M_{i,j}$ will be used for the $\left( i,j\right) $-th entry of
any matrix $M$. Thus,
\begin{equation}
A_{i,j}=x_{i-j}\cdot\operatorname*{sgn}\left( \tan\dfrac{\left( i+j\right)
\pi}{2n+1}\right)
\label{eq.darij1.1}
\tag{1}
\end{equation}
for any $i,j\in\left[ 2n\right] $.
Let $B$ be the $2n\times2n$-matrix obtained by "turning $A$ upside down",
i.e., reversing the order of the rows of $A$. Explicitly, this means that
\begin{equation}
B_{i,j}=A_{2n+1-i,j}\qquad\text{for all }i,j\in\left[ 2n\right]
.
\label{eq.darij1.2}
\tag{2}
\end{equation}
We note that $B$ can be obtained from $A$ by $n$ row-swaps (i.e., by $n$
steps, where each step swaps a pair of rows). Indeed, all we need to do is to
swap the $1$-st and the last row, then to swap the $2$-nd and the
$2$-nd-to-last row, etc., until we reach the middle of the matrix. Since each
of these swaps multiplies the determinant by $-1$, this entails that
\begin{equation}
\det B=\left( -1\right) ^{n}\det A.
\label{eq.darij1.3}
\tag{3}
\end{equation}
Now, I claim that the matrix $B$ is alternating -- i.e., that
\begin{equation}
B_{i,i}=0\qquad\text{for all }i\in\left[ 2n\right]
\label{eq.darij1.4}
\tag{4}
\end{equation}
and
\begin{equation}
B_{i,j}=-B_{j,i}\qquad\text{for all }i,j\in\left[ 2n\right]
.
\label{eq.darij1.5}
\tag{5}
\end{equation}
Indeed, in order to prove \eqref{eq.darij1.4}, it suffices to observe that
\begin{align*}
B_{i,i} & =A_{2n+1-i,i}=x_{\left( 2n+1-i\right) -i}\cdot\operatorname*{sgn}
\left( \tan\dfrac{\left( \left( 2n+1-i\right) +i\right) \pi}
{2n+1}\right) \\
& =x_{2n+1-2i}\cdot\underbrace{\operatorname*{sgn}\left( \tan\dfrac{\left(
2n+1\right) \pi}{2n+1}\right) }_{=\operatorname*{sgn}\left( \tan\pi\right)
=\operatorname*{sgn}0=0}=0.
\end{align*}
The proof of \eqref{eq.darij1.5} is not much harder (using the fact that
$\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\pi-\dfrac{\left( i-j\right)
\pi}{2n+1}$ and therefore
\begin{align}
\tan\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\tan\left( \pi-\dfrac{\left(
i-j\right) \pi}{2n+1}\right) =-\tan\dfrac{\left( i-j\right) \pi}{2n+1},
\end{align}
and furthermore $\tan$ is an odd function).
Thus, we know that the matrix $B$ is alternating. Hence, as for any
alternating $2n\times2n$-matrix, its determinant is the square of its
Pfaffian. In other words,
\begin{equation}
\det B=\left( \operatorname*{Pf}B\right) ^{2},
\label{eq.darij1.6}
\tag{6}
\end{equation}
where $\operatorname*{Pf}B$ denotes the Pfaffian of $B$. The latter Pfaffian
is a polynomial in the entries of the matrix with coefficients in $\mathbb{Z}
$. Since the entries of the matrix belong to $\mathbb{Z}\left[ \ldots
,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $, we thus conclude that the
Pfaffian belongs to $\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1}
,x_{2},\ldots\right] $ as well. In other words,
\begin{equation}
\operatorname*{Pf}B\in\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1}
,x_{2},\ldots\right] .
\label{eq.darij1.7}
\tag{7}
\end{equation}
Now, \eqref{eq.darij1.3} yields
\begin{align}
\det A=\left( -1\right) ^{n}\det B=\left( -1\right) ^{n}\left(
\operatorname*{Pf}B\right) ^{2}
\end{align}
(by \eqref{eq.darij1.6}). Because of \eqref{eq.darij1.7}, this shows that
$\det A$ equals $\left( -1\right) ^{n}\cdot P^{2}$ for some polynomial
$P\in\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right] $
(namely, for $P=\operatorname*{Pf}B$), exactly as claimed in Question 1.
In order to complete the answer to Question 1, we now need to show that each
monomial in $P=\operatorname*{Pf}B$ is of the form $x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}$ with $i_{1}+i_{2}+\cdots+i_{n}=0$. This can be done in various
ways, but the easiest is probably the following: Let us equip the polynomial
ring $\mathbb{Z}\left[ \ldots,x_{-2},x_{-1},x_{0},x_{1},x_{2},\ldots\right]
$ with a $\mathbb{Z}$-grading in which each indeterminate $x_{i}$ is
homogeneous of degree $i$. Now, recall the explicit formula for the Pfaffian
as a sum over all perfect matchings on the set $\left[ 2n\right] $ (see Definition 3 in Michel Goemans, 18.438 in Spring 2014, Lectures 4 and 6, or any
good textbook on Pfaffians). If
\begin{equation}
M=\left\{ \left\{ a_{1},b_{1}\right\} ,\left\{ a_{2},b_{2}\right\}
,\ldots,\left\{ a_{n},b_{n}\right\} \right\}
\label{eq.darij1.9o}
\tag{9}
\end{equation}
is such a perfect matching, then the corresponding addend in
$\operatorname*{Pf}B$ is
\begin{equation}
\pm B_{a_{1},b_{1}}B_{a_{2},b_{2}}\cdots B_{a_{n},b_{n}}.
\label{eq.darij1.9}
\tag{10}
\end{equation}
Each of the $n$ factors $B_{a_{i},b_{i}}$ in this product can be rewritten as
\begin{align}
B_{a_{i},b_{i}}=A_{2n+1-a_{i},b_{i}}=x_{\left( 2n+1-a_{i}\right) -b_{i}
}\cdot\left( 1\text{ or }-1\text{ or }0\right) ,
\end{align}
and thus (using our weird grading) is homogeneous of degree $\left(
2n+1-a_{i}\right) -b_{i}=2n+1-a_{i}-b_{i}$. Hence, the entire product
\eqref{eq.darij1.9} is homogeneous of degree
\begin{align*}
\sum_{i=1}^{n}\left( 2n+1-a_{i}-b_{i}\right) & =n\left( 2n+1\right)
-\underbrace{\sum_{i=1}^{n}\left( a_{i}+b_{i}\right) }
_{\substack{=1+2+\cdots+2n\\\text{(since \eqref{eq.darij1.9o} is
a}\\\text{perfect matching of }\left[ 2n\right] \text{)}}}\\
& =n\left( 2n+1\right) -\left( 1+2+\cdots+2n\right) =0.
\end{align*}
This means that this product is a $\mathbb{Z}$-linear combination of monomials
of the form $x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ with $i_{1}+i_{2}
+\cdots+i_{n}=0$. Clearly, the same must therefore holds for the polynomial
$\operatorname*{Pf}B$ (since this polynomial is a sum of such products). This
concludes the answer to Question 1.
Answering Question 2 requires proving that $\det B=1$ when all $x_{i}$ are set
to $1$. This should be easy given that $\operatorname*{sgn}\left( \tan
\dfrac{\left( i+j\right) \pi}{2n+1}\right) $ can be explicitly computed
(and the matrix $B$ becomes a circulant when all $x_{i}$ are $1$); but it's
late here and I have too many things on my list until the quarter begins. Sorry!
The answer is yes for any $n \geq 2$. (The $n=1$ case should be treated separately.)
The determinant, as a polynomial, contains $(x_m−x_{m+1})$ $(m=1,2,...,2n-1)$ and $(x_{2n}-x_1)$ as factors, because setting $x_m=x_{m+1}$ will make the $m$th and $(m+1)$th column of $\mathbb{M}_{2n}$ equal, and setting $x_{2n}=x_1$ will make the $(2n)$th and first column of $\mathbb{M}_{2n}$ equal, making the determinant zero.
Since the determinant of a skew-symmetric matrix can always be written as the square of a polynomial in the matrix entries that only depend on the size of the matrix, the determinant contains the factor $(x_1−x_2)^2(x_2−x_3)^2⋯(x_{2n−1}−x_{2n})^2(x_{2n}−x_1)^2$. As the determinant cannot have degree higher than that of this factor, the only thing left is to plug in values of $x_k$ such that LHS=RHS≠0 in the last equation stated in the question.
If we let $x_k=1$ if $k$ is odd and $0$ otherwise, then
$\mathbb{M}_{2n}(i,j)=
\begin{cases}
\phantom{-}1 \qquad \text{if $i<j$},i \text{ odd}, j \text{ even} \\
-1 \qquad \text{if $i > j$}, i \text{ even}, j \text{ odd}\\
\phantom{-}0 \qquad \text{otherwise}
\end{cases}$.
The $(2n-1)$th row has only one nonzero element, namely $\mathbb M_{2n-1,2n}=1$, and the $(2n-1)$th column has only one nonzero element, namely $\mathbb M_{2n,2n-1}=-1$. By applying the Leibniz formula for determinants, we see that $\det \mathbb M_{2n}=\det \mathbb M_{2n-2}$. By mathematical induction, we have $\det \mathbb{M}_{2n}=1$ for any $n$. As the RHS also evaluates to $1$, we claim that LHS=RHS for any $x_k$s.
Best Answer
Well, \begin{align*} \sum_{\substack{1\leq i_1<\cdots<i_{n-2}\leq n\\ \binom{n+1}2-(i_1+\cdots+i_{n-2})\,\equiv\, 1\,\, \text{mod}\, 2}}x_{i_1}\cdots x_{i_{n-2}} &=x_1\cdots x_n\cdot \left(\sum_{i+j\,\text{is odd}}(x_ix_j)^{-1}\right)\\ &=x_1\cdots x_n(x_1^{-1}+x_3^{-1}+\ldots)(x_2^{-1}+x_4^{-1}+\ldots). \end{align*}