If $a$ is odd we need at most two, because
is a factorization into parts $u_1=1$ and $u_2=a+2bi$ such that $u_1\pm u_2$ have even real and imaginary parts. This means that we can solve for $z_1$ and $z_2$ from
(using the ansatz $u_1=z_1+iz_2$, $u_2=z_1-iz_2$) as
And if $a$ is even we need at most one more, because $a-1$ is then odd.
So at this point we have $n\le3$.
It seems to me $z=2+2i$ cannot be written as a sum of two squares (study real and imaginary parts modulo four and check all the cases), meaning that $n=3$ is the answer.
Edit: Alternatively we can verify that no factorization of $2+2i=-i(1+i)^3$ leads to a factorization with both factors having the same parity in their real and imaginary parts. Working the above trick backwards then shows that it cannot be written as a sum of two squares. My testing suggests that $z=a+2bi$ is a sum of two Gaussian squares unless its real and imaginary parts are both congruent to $2$ modulo $4$.
See Gerry Myerson's answer to a newer question for more.