Does atmospheric pressure have a net contribution to this FBD (free body diagram)? 
Pressure acting on the left and right sides of the block counteract
 A: Atmospheric pressure will just act as an extra "weight" for your object, such that
$$
\left| N \right| = \left| mg \right| + \left| P_0 A_{top} \right|
$$
Think of the pressure from above as just being the weight of all the air stacked on top of the box.
From the sides you have just got
$$
\left| P_0 A_{left} \right| = \left| P_0 A_{right} \right|
$$
so this cancels and does not contribute any net force.
A: It depends on if there are air between the squared block and the floor. If there is even an small joint with air, as when a box is on the floor, atmospheric pressure is acting from all sides of the box and the downward pressure from upper side nearly balances the upward pressure from bottom side. Therefore atmospheric pressure doesn't have a net effect.
On the other hand, if we can exclude air from the bottom side, as in a suction cup, then downward pressure is unbalanced and atmospheric pressure should be taken in account in that free body diagram.
And just as a side note: In first paragraph I said that upward and downward forces nearly balance each other. In fact there is a small difference and that difference is the buoyancy of the solid in air. Anyway, such buoyancy is usually so small compared with weight that can be safely ignored unless we are dealing with very light objects - like balloons.
A: The free body diagram is accepted but one should remember to make lines of forces close to each other for forces acting on same direction.As usually it can be seen in books like HRW.
