jmc

We find that $f(1)=2^1=2.$ Then, $f(f(1))=f(2)=2^2=4$ and $f(f(f(1)))=f(4)=2^4=16.$ Therefore, $f(f(f(f(1))))=f(16)=2^{16}$ and so $\sqrt{f(f(f(f(1))))}=\sqrt{2^{16}}=2^8=\boxed{256}.$