Answer     $13$

Note that $$\log_4{(\log_4 n)}=\frac{\log_2 (\log_4 n)}{\log_2 4}=\frac{1}{2}\log_2 {(\log_4 n)}=\log_2 (log_4 n)^{\frac{1}{2}}$$

Setting this to the original equation gives $$\log_2{(\log_{16}{n})}=\log_2 {(\log_4 n)^\frac{1}{2}}\implies log_{16} n =\left( \log_4 n\right)^{\frac{1}{2}}=\left(\frac{\log_{16}n}{\log_{16}{4}}\right)^{\frac{1}{2}}=\left(2{\log_{16}n}\right)^{\frac{1}{2}}$$

Lettings $x=\log_{16} n \ne 0$ yields $$x = \left(2x\right)^{\frac{1}{2}}\implies x = 2\implies n= 256$$

Hence, the final answer is $$2+5+6=\boxed{13}$$