One can see the greatest number of cubes when looking at three
faces at once. We count the cubes on each face, giving $3\times
11^2=363$ cubes, but have to subtract from this the cubes along the
three edges that have been counted twice, and then add back for the
cube at the corner for which three faces are visible. The final
quantity is $363-(3\times 11)+1=331$ cubes.
This problem is taken from the UKMT Mathematical Challenges.