The vanishing of sum of coefficients: symmetric polynomials

Choose $n$ numbers $x_1,\dots,x_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-\ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.

If $n=2k$ is even, we substitute the roots $w_1,\dots,w_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+\ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)\dots (x-w_n)$. Then your claim reads as $$A:=\prod_{1\leqslant i<j\leqslant n} (w_i+w_j)=(-1)^{\binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first, $$ |A|^2=\prod_{i=1}^n \prod_{j\ne i,1\leqslant j\leqslant n}|w_i+w_j|=2^{-n}\prod_{i=1}^n \prod_{j=1}^n|(-w_i)-w_j|=2^{-n}\prod_{i=1}^n |f(w_i)|=\\=2^{-n}\prod_{i=1}^n\left|\frac{(-w_i)^{2k+1}+1}{-w_i+1}\right|=1, $$ since $1+(-w_i)^{2k+1}=2$ for all $i=1,2,\dots,n$ and $\prod_{i=1}^n (1-w_i)=f(1)=1$.

At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w_i+w_j$ for which $w_i$ and $w_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w_i$ and $w_j$ are complex conjugate, the sum $w_i+w_j$ is a real number whose sign is the sign of the real part of $w_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.

Here is another method using more of the theory of symmetric functions. By Enumerative Combinatorics, vol. 2, Exercise 7.30, we have $f_n(\boldsymbol{X}_n)= s_{(n-1,n-2,\dots,1)}(\boldsymbol{X}_n)$ (a Schur function). By the dual Jacobi-Trudi identity, $$ s_{(n-1,n-2,\dots,1)} = \det[e_{n-2i+j}]_{i,j=1}^{n-1},\ \ (*) $$ where $e_0=1$ and $e_k=0$ for $k<0$. Since $e_k(\boldsymbol{X}_n)=0$ for $k>n$, it follows that $\sum_\mu c_{\mu,n}$ is obtained by substituting $e_1=e_2=\cdots=e_n=1$ and $e_{n+1}=e_{n+2}=\cdots=0$ into the right-hand side of (*). When $n$ is odd, the two middle rows of the determinant are equal (in fact, they are all 1's), so the determinant is 0. If $n=2m$ then subtract row $m-1$ from row $m$, then row $m-2$ from row $m-1$, up to row 1 from row 2. Also subtract row $m+2$ from row $m+1$, row $m+3$ from $m+2$, etc. The resulting matrix $A$ can be transformed into a triangular matrix $B$ with 1's on the diagonal by row and column permutations. The permutation indexing the 1's in $A$ that become the diagonal elements of $B$ is $1,3,5,\dots,n-1,2,4,6,\dots,n-2$, which has ${m\choose 2}$ inversions, and the proof follows.