Let $\begin{cases}k=\log_2n\\U(k)=T(n)\end{cases}$ ,
Then $U(k)=2^{2^k}U(k-1)+(2^k)^{2^k}$
$U(k)=2^{2^k}U(k-1)+2^{k\times2^k}$
$U(k+1)=2^{2^{k+1}}U(k)+2^{(k+1)2^{k+1}}$
$\therefore U_c(k+1)=2^{2^{k+1}}U_c(k)$
$U_c(k)=\Theta(k)\prod\limits_k2^{2^{k+1}}$ , where $\Theta(k)$ is an arbitrary periodic function with unit period
According to http://en.wikipedia.org/wiki/Indefinite_product#Rules ,
$U_c(k)=\Theta(k)2^{\sum\limits_k2^{k+1}}$ , where $\Theta(k)$ is an arbitrary periodic function with unit period
According to http://en.wikipedia.org/wiki/Indefinite_sum#Antidifferences_of_exponential_functions ,
$U_c(k)=\Theta(k)2^{2^{k+1}}$ , where $\Theta(k)$ is an arbitrary periodic function with unit period
Let $U(k)=2^{2^{k+1}}V(k)$ ,
Then $2^{2^{k+2}}V(k+1)=2^{2^{k+1}}2^{2^{k+1}}V(k)+2^{(k+1)2^{k+1}}$
$2^{2^{k+2}}V(k+1)=2^{2^{k+2}}V(k)+2^{(k+1)2^{k+1}}$
$V(k+1)=V(k)+2^{(k-1)2^{k+1}}$
$V(k)=\Theta(k)+\sum\limits_k2^{(k-1)2^{k+1}}$ , where $\Theta(k)$ is an arbitrary periodic function with unit period
$U(k)=\Theta(k)2^{2^{k+1}}+2^{2^{k+1}}\sum\limits_k2^{(k-1)2^{k+1}}$ , where $\Theta(k)$ is an arbitrary periodic function with unit period
$T(n)=\Theta(\log_2n)4^n+4^n\left(\sum\limits_k2^{(k-1)2^{k+1}}\right)_{k\to\log_2n}$ , where $\Theta(n)$ is an arbitrary periodic function with unit period