I am not sure what is the best way to solve this problem, but here is one possible approach. We will tackle a single equation $|a^Hv|=\|g\|$ first, and deal with the system of equations later.
Given an equation $|a^Hv|=\|g\|$, let $e_1=(1,0,\ldots,0)^H$. Pick a unitary matrix $U$ whose first column is $\frac{a}{\|a\|}$ (that is, $Ue_1=\frac{a}{\|a\|}$). Any such unitary matrix will do. For instance, you may use a rotated Householder matrix, which is constructed as follows:
\begin{align*}
D&=\operatorname{diag}(e^{i\arg a_1},e^{i\arg a_2},e^{i\arg a_3},e^{i\arg a_n}),\\
x&=D^Ha-\|a\|e_1\ \in\mathbb{R}^4,\\
Q&=I-\frac{2xx^T}{\|x\|^2},\\
U&=DQ.
\end{align*}
Let $w=U^Hv$. Then the equation $|a^Hv|=\|g\|$ is equivalent to $|e_1^Hw|=\frac{\|g\|}{\|a\|}$, which is solvable by $w=(\frac{\|g\|}{\|a\|}e^{i\theta},w_2,w_3,w_4)^T$, where $\theta\in\mathbb{R}$ and $w_2,w_3,w_4\in\mathbb{C}$ are arbitrary. The vector $v$ is recovered as $v=Uw$.
Now, consider a system of equations $|\mathbf{a}_k^H\mathbf{v}|=\|\mathbf{g}_k\|$ for $k=1,2,3,4$. For each of them, the solutions are of the form $\mathbf{v}_k=U_k\mathbf{w}_k$ (here $\mathbf{w}_k$ denotes a vector; do not confuse it with vector entries). For $k<4$, define $Q_k=U_k^HU_4$. So, we are looking for $\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3,\mathbf{w}_k$ such that $\mathbf{w}_k=Q_k\mathbf{w}_4$ for each $k<4$. Note that, given any $\mathbf{w}_4$, we can always back out the 2nd to the last entries of $\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3$ from the equation $\mathbf{w}_k=Q_k\mathbf{w}_4$. So, if we can determine $\theta_1,\theta_2,\theta_3$ and $\mathbf{w}_4$, we are done. Let $\mathbf{q}_k^H$ be the first row of $Q_k$ and we perform the following partitioning:
$$
\mathbf{w}_4=\pmatrix{\frac{\|\mathbf{g}_4\|}{\|\mathbf{a}_4\|}e^{i\theta_4}\\ \widetilde{\mathbf{w}}},
\quad\pmatrix{\mathbf{q}_1^H\\ \mathbf{q}_2^H\\ \mathbf{q}_3^H}
= (\widetilde{\mathbf{q}}, \widetilde{Q}_{3\times3}).
$$
Now, put together the first rows of the system of equations $\mathbf{w}_k=Q_k\mathbf{w}_4$ for each $k$ and we get
$$
\pmatrix{\mathbf{q}_1^H\\ \mathbf{q}_2^H\\ \mathbf{q}_3^H}\mathbf{w}_4
=\pmatrix{\frac{\|\mathbf{g}_1\|}{\|\mathbf{a}_1\|}e^{i\theta_1}\\ \frac{\|\mathbf{g}_2\|}{\|\mathbf{a}_2\|}e^{i\theta_2}\\ \frac{\|\mathbf{g}_3\|}{\|\mathbf{a}_3\|}e^{i\theta_3}}
\quad\Leftrightarrow\quad
\widetilde{Q}\widetilde{\mathbf{w}}
=\pmatrix{\frac{\|\mathbf{g}_1\|}{\|\mathbf{a}_1\|}e^{i\theta_1}\\ \frac{\|\mathbf{g}_2\|}{\|\mathbf{a}_2\|}e^{i\theta_2}\\ \frac{\|\mathbf{g}_3\|}{\|\mathbf{a}_3\|}e^{i\theta_3}}
-\frac{\|\mathbf{g}_4\|}{\|\mathbf{a}_4\|}e^{i\theta_4}\widetilde{\mathbf{q}}.\tag{1}
$$
If $\widetilde{Q}$ is invertible, $(1)$ is always solvable and we can pick $\theta_1,\theta_2,\theta_3,\theta_4$ arbitrarily. However, if $\widetilde{Q}$ is singular, the system is sovlable if and only if we can choose $\theta_1,\theta_2,\theta_3,\theta_4$ such that the expression on the right end of $(1)$ lies inside the column space of $\widetilde{Q}$. I am not sure how to determine the $\theta_k$s in this case.
