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!
Best Answer
The conjecture has been proved! See the preprint Proof of five conjectures relating permanents to combinatorial sequences by Fu, Lin and me available from http://arXiv.org/abs/2109.11506.