I edited your post to make it a little easier to follow. You ask a good question:

If the vertical component of the normal force cancels out weight, and there is also a horizontal component of the normal force, then the magnitude of the normal force is *greater* than the magnitude of the weight - what provides the extra normal force?

First, why do you assume that the entirety of the normal force must be equal to weight? The normal force describes the interaction between the car and the road, and since you are accelerating (circular motion), there are certainly more forces at play than just the weight.

Since you are on an angle, then the road "pushes" a little in the horizontal direction, and a little in the vertical direction. The net force pointing in the direction of the center of the circular path is by definition the centripetal force. Since the only object the car interacts with (other than Earth) is the banked road, then that gives us [3]. Since we know there is an interaction between car and Earth (weight), and that the car is not moving in the horizontal direction, then we know [1] must be true. Equation [1] isn't necessarily an assumption. Its true based on the conditions specified.

To answer your question, the "extra" normal force is the centripetal force.

Imagine a book sitting on a table. The book weights 10 N. You also apply 10 N of force in the down direction with your hand. What is the normal force (interaction between desk and book)? The answer is 20 N. The force your hand applies to the book contributes to the interaction between the desk and book. The same is true with the car and track. Yes, you have the weight of the car "pushing" on the track, but you also have the centripetal force caused by the car's acceleration.