Biquadratic modulus

Since $f(x)$ has all real roots, say these are $p,q,r,s$ so that $f(x)=(x-p)(x-q)(x-r)(x-s)$

In particular, $f(i)=(i-p)(i-q)(i-r)(i-s)$

and

$f(-i)=(i+p)(i+q)(i+r)(i+s)$

Since these are conjugates, $|f(i)|^2=f(i)f(-i)=\left(p^2+1\right)\left(q^2+1\right)\left(r^2+1\right)\left(s^2+1\right)$

Each of these brackets is at least one; but $|f(i)|=1$ , so we must have $p=q=r=s=0$

So $f(x)=x^4$ and $f(0)+f(2)+f(4)=\boxed{272}$ .