Note that $ z^2 + z + 1 = 0 $ and $ z^3 = 1 $. Also, note that $ s_2 = zs_1 $ and $ s_3 = z^2 s_1 $. The sum of the coefficients of $ g(x) $ is \begin{align*} g(1) &= (1 - s_1)(1 - s_2)(1 - s_3) \\ &= (1 - s_1)(1 - s_1 z)(1 - s_1 z^2) \\ &= 1 - (1 + z + z^2) s_1 + (z + z^2 + z^3) s_1^2 - z^3 s_1^3 \\ &= 1 - s_1^3.\end{align*}We have that \[s_1^3 = r_1^3 + r_2^3 + r_3^3 + 3r_1^2 r_2 z + 3r_1^2 r_3 z^2 + 3r_2^2 r_3 z + 3r_2^2 r_1 z^2 + 3r_3^2 r_1 z + 3r_3^2 r_2 z^2 + 6r_1 r_2 r_3.\]Note that $ r_1, $ $ r_2, $ and $ r_3 $ are all real, and the real part of both $ z $ and $ z^2 $ are $ -\frac{1}{2}, $ so the real part of $ s_1^3 $ is \begin{align*} &r_1^3 + r_2^3 + r_3^3 - \frac{3}{2} (r_1^2 r_2 + r_1 r_2^2 + r_1^2 r_3 + r_1 r_3^2 + r_2^2 r_3 + r_2 r_3^2) + 6r_1 r_2 r_3 \\ &= (r_1 + r_2 + r_3)^3 - \frac{9}{2} (r_1 + r_2 + r_3)(r_1 r_2 + r_1 r_3 + r_2 r_3) + \frac{27}{2} r_1 r_2 r_3 \\ &=3^3 - \frac{9}{2} (3)(-4) + \frac{27}{2} (-4) = 27.\end{align*}Hence, the real part of the sum of the coefficients of $ g(x) $ is $ 1 - 27 = \boxed{-26} $.