Tidal forces come from gradients in gravitational fields.
Along the horizontal axis of the station, the distance to Earth stays the same, so there's no gradient and therefore no tidal forces.
To find the gradient along the vertical axis, we can simply use the derivative.
$$\left(\frac{\mu}{r^2}\right)' = -\frac{2\mu}{r^3}$$
At the the orbital radius of the ISS, this works out to a gradient of $2.6 \cdot 10^{-6} s^{-2}$
The ISS is quite flat, so you can only multiply that gradient by the handful of metres between the "floor" and "ceiling" in the modules, so in the order of $\approx10^{-5} m/s^2$
That's still about a magnitude more than the acceleration due to aerodynamic drag on the station, $\approx 10^{-6} m/s²$
It's not the dominating acceleration within the station though. Lost items end up in the air filters.
edit: thanks, uhoh, for reminding me that there's a third axis to consider, the north-south axis. Rigidity here should provide a slight effect, lower than at the vertical axis, but it isn't cancelled out by orbital velocity in the same way the prograde-retrograde axis is. Here's a NASA page mapping out all the 3 axis