We look for solutions $ a\ge b\ge c\ge d$. $ n=0$ gives only solution $ (2,1,1,1)$. $ n=1$ gives solutions $ (5,1,1,1)$, $ (4,2,2,2)$, $ (3,3,3,1)$. Take now $ n\ge2$. Then $ a^2+b^2+c^2+d^2=7\cdot4^n=8k$. $ x^2$ mod $ 8\in\{0,1,4\}$ so it's easy to see that all $ a,b,c,d$ must be even. Let $ a=2a'$, $ b=2b'$, $ c=2c'$, $ d=2d'$. Then $ a'^2+b'^2+c'^2+d'^2=7\cdot 4^{n-1}$. Prooceding just as before we finally obtain the exponent of $ 4$ less than $ 2$. It easily follows that $ (a,b,c,d)=(5\cdot2^{n-1},2^{n-1},2^{n-1},2^{n-1})$ or $ (a,b,c,d)=(2^{n+1},2^n,2^n,2^n)$ or $ (a,b,c,d)=(3\cdot2^{n-1},3\cdot2^{n-1},3\cdot2^{n-1},2^{n-1})$. Disregarding $ a\ge b\ge c\ge d$ we obtain all solutions.

An easier version, but which holds the same tools to solve the above: Solve $x^2+y^2=3^{2014}$ in nonnegative integers.