Is there a fundamental lower bound to resistor precision? The resistor noise index provides a lower bound to how precise a resistor can be measured, even using ideal instrumentation. Effectively, it even makes no sense to define the resistance any more precise than given by its noise index. Intuitively, this can be understood as more probe charge (either due to higher current or longer integration time) also influences the resistor more.
However, by increasing the volume of the resistor, one can reduce the noise index: Below in Fig. 1, both resistances are equal to $R$, but the right assembly uses four identical resistors each contributing uncorrelated noise. It can be shown, that the 4 resistor assembly has a twice lower total excess noise and thus lower noise index. Therefore, the resistance of the right side assembly can be defined twice more precise than the single resistor on the left side. By quadrupling the volume of the resistor and leaving everything else unchanged, the excess noise drops to half demonstrating the relation $n\propto V^{-1/2}$.
My question is:
How far can this be taken ? One could always add more resistors achieving extremely precise assemblies. But fundamentally, to probe the resistance electrically one relies on electron-electron interaction which are quantum mechanical processes. So I figure that the uncertainty principle might sneak into the backyard somehow.

Figure 1: Using 4 identical resistors, one can make an assembly with the same resistance, but twice lower excess noise
 A: I am not sure if this is a way to have more precise assemblies in practice.

*

*The excess noise is material dependent so choose different materials, so in practice using a metal wire resister rather than a carbon thin film or other type might be better.

*As you increase the complexity of the circuit you are also increasing the number of contacts to the resistors as well as creating parasitics inductance and capacitance so that might be limiting.

*You start with one resistor, then 4, then 16, then 48 resistors as you try to extend the method... and the current going through an individual resistor is 1/2, then 1/4, then 1/8 and corresponding voltage drops.

*Your assumption is that the network of resistors are all identical, but each would also have some variation.

I am not sure if the excess noise is a measure of the precision value of the resistor, but for a constant current source it would be the mean of the voltage divided by the current.
Edited in response to comment: Take a look at the discussion in Belev which is a paper about measuring excess noise in resistor networks. It compares various types (materials) and some other issues.

This finding was unexpected as it is contradictory to the usual assumption that for a given resistor type NI increases with resistance as the volume of the resistive element decreases 1 [2] [4]. It could be indicative of dominant noise mechanisms originating outside the film, e.g. in the contacts [14] [22] or in points of high local current density such as trim links [23].

It then goes on to discuss some of the limitations.
In terms of fundamental physical limits, I think the answer still depends on the fabrication, in particular with laser trimming of resistors a localized changes in structure and cracking, as well as limitations due scattering and other issues which can be termed variations in mobility. Similarly as the Belev paper points out, surface roughness matters a great deal.
You can perhaps argue surface roughness and fabrication issues are not the fundamental physical limitations. But they are directly related to charge trapping, scattering of a variety of types and other considerations of electron transport. They can be very appreciable as the films get thin compared to the bulk.
I don't think you need to bring electron confinement, or quantum mechanics into it for any reasonable values of "resistors". You can easily see that electron conductance is quantized with a relay and a fast oscilloscope where as the relay is tapped you form a resistor wire only a few atoms in diameter. So for very small resistors at the nanoscale it is a whole other ball game.
If you go the other extreme and assume you continue build out your network such that the current is decreased for each resistor with reasonable resistance in the network, I suppose eventually you could argue that you would only have a few electrons going through each resistor but I don't see how that brings quantum effects into it other than it becoming more of a statistical problem.
A: Resistor excess noise is poorly understood. There is, as far as I know, no verified model that can yield the answer you seek.
This is a fine example of a physical effect that is observable with simple lab equipment but remains a mystery.
