An arbitrary surface encloses a dipole.

From Gauss' law, the electric flux through an enclosed surface is given by

$\int_{S} E \cdot d s=\frac{q}{\varepsilon_{o}}$

Where, q is the net charge enclose by the surface

Now total charge enclose by the dipole = (‒ q + q) = 0

Therefore, electric flux through a surface enclosing a dipole = 0.