smc

The domain of $f_1(x)=\sqrt{1-x}$ is defined when $x\le 1$. $$f_2(x)=f_1\left(\sqrt{4-x}\right)=\sqrt{1-\sqrt{4-x}}$$ Applying the domain of $f_1(x)$ and the fact that square roots must be positive, we get $0\le \sqrt{4-x}\le 1$. Simplifying, the domain of $f_2(x)$ becomes $3\le x\le 4$. Repeat this process for $f_3(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}$ to get a domain of $-7\le x\le 0$. For $f_4(x)$, since square roots must be nonnegative, we can see that the negative values of the previous domain will not work, so $\sqrt{16-x}=0$. Thus we now arrive at $16$ being the only number in the of domain of $f_{4}x$ that defines $x$. However, since we are looking for the largest value for $n$ for which the domain of $f_n$ is nonempty, we must continue checking until we arrive at a domain that is empty. We continue with $f_5(x)$ to get a domain of $\sqrt{25-x}=16⟹x=-231$. Since square roots cannot be negative, this is the last nonempty domain. We add to get $5-231=\boxed{-226}$.