I doubt it. Let's look at $m_{(1,0,0,0,0)}$. You want the solutions of $a+b+c+d+e=0$ with all variables on the unit circle. It will be hard enough to find a formula for that special case, much less for the general case.
Note that $m_{(1,0)}$ is $a+b=0$ which is solved by $a=e^{it}$, $b=e^{i(t+\pi)}$. Then $m_{(1,0,0)}$ is $a+b+c=0$, which forces $a,b,c$ to be vertices of an equilateral triangle. Next, $m_{(1,0,0,0)}$ is $a+b+c+d=0$, and with a bit of work you can show that $a,b,c,d$ must be vertices of a rectangle. But once you get up to 5 unknowns the geometric argument doesn't give you anything that simple.