I assume you want to keep the beans suspended in the air, like they do with persons at skydive simulators.
In order to do that, you do need a high speed airflow (or CFM). The (static) pressure that the fan can maintain doesn't matter here. The high speed airflow creates dynamic pressure that acts on the beans and keeps them suspended, which is what your looking for if I understand correctly. At terminal velocity, the force that acts on the object due to drag, is equal to the force that pulls it down, which we call gravity.
Consider the following formula for the terminal velocity of an object:
$v=\sqrt(2mg/\rho*C_d*A)$
Where $v$ is the velocity on an object, relative to the medium/stream in metres per second.
$m$ is the mass of the object(beans) in kilogram.
$g$ is the gravitational acceleration, which is about $9.81m/s^2$ everywhere on Earth.
$\rho$ is the density of the medium, which is 1.2kg per cubic metre for air at standard conditions.
$C_d$ is the drag coefficient of the object. This is a factor usually between near 0 and 1. It's defined by the shape and skin texture of the object. A sphere has a $C_d$ of about 0.9, so i'd expect a bean to have a $C_d$ of about 0.85. It depends which orientation the bean has relative to the airflow.
$A$ is the frontal area of the object. This also depends on the orientation of the bean.
Since we can't change the properties of the beans, nor can we significantly change the density of the air, all we're left with is the velocity as tunable parameter.
I can't find the properties of a bean in my physics book, but let's suppose it's as dense as average wood: about 700kg/m3, and that the bean has the shape of a sphere with a diametre of 8mm. (about 1/3rd of an inch) Then the frontal area of the bean will be:
$A=1/4*\pi*D^2=1/4*\pi*8^2=50.3mm^2$ or 5.03e-5 m2.
The volume of the spherical bean will be:
$V=1/6*\pi*D^3=268mm^3$ or 2.68e-7 m3.
Which at a density of 700kg/m3, gives us a mass of the bean of:
$m=\rho*V=700*2.68e-7=0.188grams$ or 1.88e-4 kg.
If we fill in these values in the first formula, it gives us a velocity of:
$\sqrt((2*1.88e-4*9.81)/(1.2*0.85*5.03e-5))=8.5m/s$
That is the needed velocity of the air to keep them suspended. If you multiply it by the surface of your fan, you get the needed airflow. This is assumed the surface of the airduct above the fan is the same surface as the fan. Your 120mm fan has a surface of:
$A=1/4*\pi*D^2=1/4*\pi*0.12^2=0.0113m^2$
Which means that the volume air flow must be:
$Volumeflow=v*A=8.5*0.0113=0.096m^3/s$ or 96L/s or 203CFM.
I'd choose a fan that has a higher CFM rating(300CFM or so), to create a margin to make up for any errors made by assumptions in these calculations. You can always easily throttle down the fan. And also, your mesh, the heater, the beans, and any other construction will restrict the airflow, which makes it impossible for the fan to provide advertised CFM rating. That rating is probably based on an unrestricted fan. That's why you need to aim higher for CFM.
If you want to warm up the beans, i'd recommend to create a construction that makes the fan recirculate the air. So create your airduct like a donut or a square donut. Not only does it allow you to gradually heat up the air and the beans without losing your heat, it also keeps the load of the fan at a minimum.