Determinant of symmetric Latin square Let $n=2m$ be an even number. Let us construct $n\times n$ symmetric matrices $S_n$ in the following way. The entries are indeterminates $X_1,\ldots,X_{n-1}$. We choose a $1$-factorization of the complete graph over $[\![1,n]\!]$ (which exists because $n$ is even). The $1$-factors (these are perfect matchings) are labelled $F_1,\ldots,F_{n-1}$. The indeterminate $X_r$ is attributed to the $(ij)$-entry whenever the edge $(ij)$ belongs to $F_r$. On the diagonal, we put zeros. In other words, every line and every column of $S$ contains a zero and all the indeterminates exactly once. You may call this a symmetric Latin square. For instance
$$S_2=\begin{pmatrix} 0 & X \\ X & 0 \end{pmatrix},\qquad S_4=\begin{pmatrix} 0 & X & Y & Z \\ X & 0 & Z & Y \\ Y & Z & 0 & X \\ Z & Y & X & 0 \end{pmatrix}.$$
It seems that the determinant $D_n$ of $S_n$, a homogeneous polynomial in the indeterminates, is symmetric.
One verifies easily that for $n=2$ or $4$, $D_n$ splits:
$$D_2=-X^2,\qquad D_4=(X+Y+Z)(X-Y-Z)(Y-Z-X)(Z-X-Y).$$ 
Actually, the characteristic polynomials $P_n(T,X_1,\ldots,X_{n-1})$ split for $n=2$ or $4$ :
$$P_2=(T-X)(T+X),\qquad P_4=(T-X-Y-Z)(T+X+Y-Z)(T+Y+Z-X)(T+Z+X-Y).$$ 

Is it true that $D_n$ or $P_n$ always split into linear factors ?

Notice that $P_n$ has always the factor $T-X_1-\cdots-X_{n-1}$.
For $n\ge6$, the $1$-factorization is not unique. There are already 6240 of them for $n=8$. Does the answer depend on the choice of the $1$-factorization ?
Edit. Patrik's negative answer raises a far-reaching question. The original question ressembles vaguely that of the splitting of the determinant associated with the Cayley graph of a finite group $G$. In the latter case, we know that the determinant splits into linear factors if and only if $G$ is abelian. For a general group, the nature of the splitting is given by Frobenius' Theorem about the irreducible characters of $G$. In the present situation, I suspect that another algebraic theory will tell us what are the degrees of the irreducible factors of $P_n$. When $n$ is small, these degrees are small, because the underlying algebra must be trivial ($n=2$ or $4$) or very simple ($n=8$).
 A: No it's not true in general. I claim that the following matrix
\begin{bmatrix} 
t&x&y&z&w&u&v&r\\
x&t&z&y&r&w&u&v\\
y&z&t&x&v&r&w&u\\
z&y&x&t&u&v&r&w\\
w&r&v&u&t&x&y&z\\
u&w&r&v&x&t&z&y\\
v&u&w&r&y&z&t&x\\
r&v&u&w&z&y&x&t
\end{bmatrix}
is of the desired form (when $t=0$) and has determinant
$[(r-u)^2-(t-y)^2+(v - w + x - z)^2]\cdot $
$[(r-u)^2-(t-y)^2+(v - w - x + z)^2]\cdot$ 
$(r + u + t + y + v + w + x + z)
(r + u  -t - y + v + w - x - z)$ 
$(r + u + t + y - v - w - x - z)(r + u -t - y - v - w + x + z)$.
Realize $K_8$ as two copies of $K_4$ and connect all vertices (one with vertex set $\{1,2,3,4\}$ and one with vertex set $\{1',2',3',4'\}$). The 1-factors $x,y,z$ come from uniting corresponding matchings of $K_4$ and $K_4'$, and the edges between $K_4$ and $K_4'$ are decomposed into the 1-factors $w,u,v,r$ given by $\{11',22',33',44'\},\{12',23',34',41'\},\{13',24',31',42'\},\{14',21',32',43'\}$ respectively, completing the 1-factorization. The determinant computation was done using a computer, so unfortunately I have no particular understanding of what is going on.
