<b>EDITED.</b> 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 <a href="http://oeis.org/A034807">Lucas polynomial</a>.
I think it should be possible to prove these formulas using the <a href="https://mathoverflow.net/questions/440962/a-question-on-the-real-root-of-a-polynomial/441115#comment1137548_440975">recurrence</a> $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in <a href="https://mathoverflow.net/questions/440962/a-question-on-the-real-root-of-a-polynomial/441115#comment1137991_441115">Ira Gessel's comment below</a>, 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.