smc

We can use the fact that ${\log }_{a^b}c=\frac{1}{b}{\log }_{a}c.$ This can be proved by using Change of Base Formula to base $a.$ So, the original equation ${\log }_2({\log }_{2^4}n)={\log }_{2^2}({\log }_{2^2}n)$ becomes $${\log }_2\left(\frac{1}{4}{\log }_{2}n\right)=\frac{1}{2}{\log }_2\left(\frac{1}{2}{\log }_{2}n\right).$$ Using log property of addition, we expand both sides and then simplify: $$\begin{array}{rl}{\log }_2\frac{1}{4}+{\log }_2({\log }_{2}n)& =\frac{1}{2}\left[{\log }_2\frac{1}{2}+{\log }_2({\log }_{2}n)\right]\\ {\log }_2\frac{1}{4}+{\log }_2({\log }_{2}n)& =\frac{1}{2}\left[-1+{\log }_2({\log }_{2}n)\right]\\ -2+{\log }_2({\log }_{2}n)& =-\frac{1}{2}+\frac{1}{2}({\log }_2({\log }_{2}n)).\end{array}$$ Subtracting $\frac{1}{2}({\log }_2({\log }_{2}n))$ from both sides and adding $2$ to both sides gives us $$\frac{1}{2}({\log }_2({\log }_{2}n))=\frac{3}{2}.$$ Multiplying by $2,$ exponentiating, and simplifying gives us $$\begin{array}{rl}{\log }_2({\log }_{2}n)& =3\\ {\log }_{2}n& =8\\ n& =256.\end{array}$$ Adding the digits together, we have $2+5+6=\boxed{13}.$