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EDITED. First let us define $$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$ Then empirically, $$\eqalign{f_{2n}(x) &= g_n(x)^2\cr f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_{n+2}(x)\cr}$$ where $h_n(x)$ is a Lucas polynomial. I think it should be possible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired properties of the roots of $f_{2n}(x)$ and $f_{4n+1}(x)$ follow from the properties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.