As at a corner, 8 cubes can be placed symmetrically. flux linked with each cube (due to charge Q at the $\text{corner) will be}\frac{Q}{8{\epsilon }_0}\text{.}$
Now for the faces passing through the edge A, electric field E at a face will be parallel to area of face and so flux for these faces will be zero. Now as the cube has six faces and flux linked with three faces(through A) is zero, so flux linked with 
remaining three faces will be $\frac{Q}{8{\epsilon }_0}$ 

Hence, electric flux passed through all the six faces of the cube is $\frac{Q}{8{\epsilon }_0}$