Then why does the rule single out measurements?
There is no special interaction associated with measurements in classical theory. Look, try to formulate the Born rule without mentioning "measurement" or a macroscopic/microscopic distinction. It is not possible. In contrast, the rules for classical mechanics can be formulated without mentioning measurement. That doesn't mean that there are no measurements in classical mechanics, but that measurements are a derived concept, not a primitive concept.
We could try to make measurements a derived concept in QM, as well.
What is a measurement? Well, a first cut at this is that a measurement is an interaction between one system (the system to be measured) and a second system (the measuring device) so that the quantity to be measured causes a macroscopic change in the measuring device. Roughly speaking, we have:
- A system being measured that has some operator [itex]O[/itex] with eigenvalues [itex]o_i[/itex].
- A measuring device, which has macroscopic states "[itex]S_{ready}[/itex]" (for the state in which it has not yet measured anything) and "[itex]S_i[/itex]" (for the state of having measured value [itex]o_i[/itex]; each macroscopic state will correspond to a huge number of microscopic states.)
- The device is metastable when in the "ready" state, meaning that a small perturbation away from the ready state will lead to an irreversible, entropy-increasing transition to one of the states [itex]S_i[/itex].
- The interaction between system and measuring device is such that if the system is in a state having eigenvalue [itex]o_i[/itex], then the device is overwhelmingly more likely to end up in the state [itex]S_i[/itex] than any other macroscopic state [itex]S_j[/itex]. (There might be other macroscopic states representing a failed or ambiguous measurement, but I'll ignore those for simplicity.)
Note: There is a bit of circularity here, in that to really make sense of the Born rule, I have to define what a measuring device is, and to define a measuring device, I have to refer to probability (which transitions are much more likely than which others), which in quantum mechanics has to involve the Born rule. What we can do, though, is treat the fact that the device makes a transition to state [itex]S_i[/itex] when it interacts with a system in a pure state with eigenvalue [itex]o_i[/itex] as initially being a matter of empirical observation, or it can be derivable from classical or semiclassical physics.
Then the Born rule implies that if the system to be measured is initially in a superposition of states of the form: [itex]\sum_i c_i |\psi_i\rangle[/itex] where [itex]|\psi_i\rangle[/itex] is an eigenstate of [itex]O[/itex] with eigenvalue [itex]o_i[/itex], then the measuring device will, upon interacting with the system, make a transition to one of the macroscopic states [itex]S_i[/itex] with a probability given by [itex]|c_i|^2[/itex].
Now, at this point, I think we can see where the talk about "measurement" in quantum mechanics is something of a red herring. Presumably, the fact that a pure eigenstate [itex]|\psi_i\rangle[/itex] of the system triggers the measuring device to go into macroscopic state [itex]S_i[/itex] is in principle derivable from applying Schrödinger's equation to the composite system, if we know the interaction Hamiltonian. But because of the linearity of quantum evolution, one would expect that if the initial state of the system were a superposition [itex]\sum_i c_i |\psi_i\rangle[/itex], then the final state of the measuring device would be a superposition of different macroscopic states, as well. (Okay, decoherence will actually prevent the device from being in a superposition, but the combination system + device + environment would be in a superposition.) So the Born rule for measurements can be re-expressed in what I think is an equivalent form that doesn't involve measurements at all:
If a composite system is in a state of the form [itex]|\Psi\rangle = \sum_i c_i |\Psi_i\rangle[/itex], where for [itex]i \neq j[/itex], [itex]|\Psi_i\rangle[/itex] and [itex]|\Psi_j\rangle[/itex] represent macroscopically distinguishable states, then that means that the system is in exactly one of the states [itex]|\Psi_i\rangle[/itex], with probability [itex]|c_i|^2[/itex]
Note the difference with the usual statement of the Born rule: I'm not saying that the system will be measured to have some eigenvalue [itex]\lambda_i[/itex] with probability [itex]|c_i|^2[/itex], because that would lead to an infinite regress. You would need a microscopic system, a measuring device to measure the state of the microscopic system, a second measuring device to measure the state of the first measuring device, etc. The infinite regress must stop at some point where something simply HAS some value, not "is measured to have some value".
But with this reformulation of Born's rule, which I'm pretty sure is equivalent to the original, you can see that the macroscopic/microscopic distinction has to be there. You can't apply the rule without the restriction that [itex]i \neq j[/itex] implies [itex]|\Psi_i\rangle[/itex] is macroscopically distinguishable from [itex]|\Psi_j\rangle[/itex]. To see this, take the simple case of a single spin-1/2 particle, where we only consider spin degrees of freedom. A general state can be written as [itex]|\psi\rangle = \alpha |u\rangle + \beta |d\rangle[/itex]. Does that mean that the particle is "really" in the state [itex]|u\rangle[/itex] or state [itex]|d\rangle[/itex], we just don't know which? No, it doesn't mean that. The state [itex]|\psi\rangle[/itex] is a different state from either [itex]|u\rangle[/itex] or [itex]|d\rangle[/itex], and it has observably different behavior.
If you eliminate "measurement" as a primitive, then you can't apply the Born rule without making a macroscopic/microscopic distinction.