Sorry for the late response. Here's a proof of the open mapping theorem assuming the maximum modulus principle.
First, we need the "minimum modulus principle". That is, if $f$ is a non-constant analytic function on an open connected set $D \subset \mathbb{C}$, and $f$ has no zeroes in $D$, then $| f |$ cannot attain a minimum in $D$. The proof follows trivially by applying the maximum modulus principal to the function $1/f$ which is analytic on $D$.
Now suppose $D \subset \mathbb{C}$ is open and connected, and $f$ is a non-constant analytic function on $D$. Let $U \subset D$ be open, and let $w_0 \in f(U)$, say $w_0=f(z_0)$ with $z_0 \in U$. We must show that there is a disc centered at $w_0$ which is contained in $f(U)$.
Choose $t>0$ so that $\overline {D_t(z_0)} \subset U$ and $f(z) \neq w_0$ for any $z \in \overline {D_t(z_0)}$ other than $z_0$. Let $m=\inf \{|f(z)-w_0| : |z-z_0|=t \} > 0 $. Suppose $|w-w_0| < m/3$, and that there is no $z \in U$ such that $f(z)=w$. Then the function $g(z)=f(z)-w$ is analytic, non-constant, and has no zeroes in the open connected set $D_t(z_0)$, so the minimum modulus principle shows that $g$ cannot attain a minimum modulus in $D_t(z_0)$. However, $g$ does attain a minimum modulus in the compact set $ \overline {D_t(z_0)} $, so this minimum modulus must occur on the boundary circle defined by $|z-z_0|=t$. But if $|z-z_0|=t$, then
$|g(z)|=|f(z)-w| \geq |f(z)-w_0| - |w_0-w| \geq 2m/3 $, and
$|g(z_0)| = |w_0-w| < m/3 < 2m/3$.
This gives a contradiction since $z_0$ is obviously in the interior of the disc in question. Therefore $f(z)=w$ for some $z \in D_t(z_0)$, and $D_{m/3}(w) \subset f(U)$, showing that $f(U)$ is open and proving the theorem.
Using the maximum principle on $f$ is of course allowed because the hypothesis are satisfied. But this is not very concluding : the only information we know is that $u^2 + v^2$ take its maximum on $\partial D$ .
Using $g$ is more appropriate : $|g| = |e^u|$ and therefore $g$ verifie the Maximum Modulus Principle $\Leftrightarrow$ $u$ verifies the MMP too.
But in fact it's true for $v$ too : take $f' = if$ and $g' = e^{f'}$. Because $g'$ is analytic, the maximum principle applies, and by the same argument (namely that $|g'| = |v|$) you can conclude that $v$ also take its maximum on $\partial D$ (until it's constant).
Best Answer
I will give a proof for $n=2$ because I have it typed out already. The same argument works for any $n$: given $f$ and $g$ analytic such that the maximum of $|f|+|g|$ on $\overline {U}$ is attained at an interior point $a$ we have $\left\vert f(a)\right\vert +\left\vert g(a)\right\vert \geq \left\vert f(z)\right\vert +\left\vert g(z)\right\vert $ $\forall z\in \Omega .$ Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where $s$ and $t$ are chosen such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to $[0,\infty ).$ This reduces the proof to the case when $f(a)$ and $g(a)$ both belong to $% [0,\infty ).$ We now have $$f(a)+g(a)\geq \left\vert f(z)\right\vert +\left\vert g(z)\right\vert \geq |f(z)+g(z)|$$ Maximum Modulus principle applied to $f+g$ shows that $f+g$ is a constant. Now $$f(a)+g(a)\geq \left\vert f(z)\right\vert +\left\vert g(z)\right\vert \geq \Re f(z)+\Re g(z)=\Re(f(z)+g(z))$$ $$=\Re (f(a)+g(a))$$ which implies that equality holds throughout. In particular $% \left\vert f(z)\right\vert =\Re(f(z))$ and $\left\vert g(z)\right\vert =\Re(g(z))$ for all z]. Hence $f$ and $g$ are both constants (because their imaginary parts vanish) in which case there is noting to prove.