According to Gauss's Law, the total electric flux $\Phi_E$ through a closed surface is equal to the net charge enclosed divided by $\epsilon_0$ ($\Phi_E = \frac{q_{\text{enclosed}}}{\epsilon_0}$). An electric dipole consists of two equal and opposite charges ($+q$ and $-q$) separated by a small distance, so the net charge of a single dipole is zero ($q_{net} = +e - e = 0$). Since there are eight dipoles inside the cube, the total enclosed charge is $8 \times 0 = 0$. Consequently, the total electric flux coming out of the cube is zero.