You cannot simply divide by 4, no. If you set up that formula for each wheel, then you'll have to take into account that the speeds and distances are different.
The one thing all four wheels have in common is angular velocity $\omega$. Even at a 90 degree turn, the rear and front wheels spin equally fast (degrees per second) around the rotation point. Otherwise the car would be breaking apart.
The angular velocity relation $$v=\omega r$$ helps you calculate the linear speed $v$ for each individual wheel, since you know the distances $r$. With this $v$ and $r$ per wheel you can calculate the centripetal acceleration for each wheel.
I would then divide the mass by 4. That would be a necessary assumption, namely that each wheel "carries" equally much mass. Then the centripetal force on each wheel can be calculated.
I have not done the calculations, but I would expect all four forces to be different. And varying differently with the turning angle of the front wheels (some will be cosine and some sine to the angle). For a 90 degree turn you will have a rotation point located at the first rear wheel. That wheel spins around, but doesn't move linearly. It has zero speed $v$ and distance $r$. According to your formula (if you plug in the relation $r=v/\omega$ in place of $r$), zero speed indeed shows zero centripetal force.
Is it also safe to assume the forces on the front wheels are equal to eachother, and also the same for the rear wheels?
If the turn is big, then differences in the distance $r$ become negligible. The differences in speeds will as well become negligible. And then all forces are more or less equal, yes.