For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that:

$\log_b a$ is defined if and only if $a>0.$
For $0b^c.$
$\log_b a>c$ if and only if $01,$ we conclude that:

$\log_b ac$ if and only if $a>b^c.$

Therefore, we have
$$\begin{align*} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\ &\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\ &\implies 1<\log_{\frac1{16}}x<2 \\ &\implies \frac{1}{256}<x<\frac{1}{16}. \end{align*}$$
The domain of $f(x)$ is an interval of length $\frac{1}{16}-\frac{1}{256}=\frac{15}{256},$ from which the answer is $15+256=\boxed{\textbf{(C) }271}.$
Remark
This problem is quite similar to 2004 AMC 12A Problem 16.
~MRENTHUSIASM