Bulgaria

Note that if $x_1,\dots ,x_{2^n}$ are the zeroes of $f_n$, then the zeroes of $f_{n+1}$ are the solutions of the equations $f_1(x)=x_k$, $1\le k\le 2^n$. Moreover, the equation $f_1(x)=a$ has two different real roots if and only if $a>m:=\min f_1$.

Assume that $f_2$ has four non-positive different zeroes. Then it is easy to see that $f_1$ has two zeroes $x_1<x_2\le 0$ and $x_1>m$. Using the argument from the beginning, it follows by induction on $n$ that all the zeroes of $f_{n+1}$ are different and belong to the interval $(x_1,x_2]$.