This is another comment. Empirically,
$$f_{2n}(x) = \biggl(\sum_{k=0}^{n/2} {n-k\choose k} x^{n-k}\biggr)^2.$$
From this, it should be easier to prove the desired facts about the roots. I think one should be able to guess a similar expression for $f_{2n+1}(x)$ and prove it using the recurrence.