While the block is falling, its acceleration is due to the gravitational force provided by the Earth, which is, in modulus, $F = mg = 9.81 \cdot 10^3 \text{N}$, about $10\text{kN}$. Newton's 3rd law applied to this force tells you that the block is pulling the Earth "up" with the very same force.
I assume that when you say
"10 N due to gravity"
you actually mean $10\text{kN}$.
This is not the force acting between the block and any of the molecules of the surrounding air.
During fall (but also at rest, for what matters) molecules collide with the block: during collision, the block will exert a force on each molecule which is identical, and opposite in direction, to the force that each molecule exerts on the block. However, these forces have nothing to do with the gravitational force acting on the block: it is another force, ultimately due to the electrostatic interaction between the molecule and the block during collision, which is quite complex.
Considering a single molecule, you might in principle calculate how large is this force by considering the initial (just before collision) momentum and kinetic energy of the block and of the molecule and by applying the conservation laws for these two quantities (to be fair, energy won't be exactly conserved). The point is that, once you have the final momenta, you could compute the force, but you would need to know how long the collision lasted - and you should assume that the force is constant!
Finally, as other answers already pointed out, when you put together many molecules, the forces due to collisions in one direction are on average balanced out by those in the opposite direction if the block is at rest. If it is moving, collision "from below" will overcome those "from above", and you get a very naive but intuitive model of viscous drag.