Let’s say that Alice, Bob, and Charlie hold random bits, which are either all or all (so, they’re classically correlated). If all three of them get together, they can see that their bits are correlated, and the same is true if only two of them are together.
If Alice, Bob, and Charlie all have either the bits $\{1_A, 1_B, 1_C\}$ or the bits $\{0_A, 0_B, 0_C\}$, upon multiple measurements, Bob and Charlie will notice that their bits are always the same as Alice's bits. This is the classical "correlation" Aaronson mentions.
But now suppose the three players share a GHZ state. With all three of them, they can see that the state is entangled (...)
A GHZ state allows for non-classical (or quantum) correlations. If Alice, Bob, and Charlie are together, then with some measurements they can detect some non-trivial correlations; this concept is well explained here.
but what if Charlie is gone? Can Alice and Bob see that they’re entangled with each other? No. To see this, observe that by the No-Communication Theorem, Charlie could’ve measured without Alice and Bob knowing.
If Charlie measured his qubit then depending on whether he gets $|0\rangle_C$ or $|1\rangle_C$, the remaining composite state of Alice and Bob's qubits would be either $|00\rangle_{AB}$ or $|11\rangle_{AB}$ respectively. This is clear from the definition of the GHZ state i.e.,
$$|\mathrm{GHZ}_3\rangle = \frac{|00\rangle_{AB} |0\rangle_C + |11\rangle_{AB}|1\rangle_C}{\sqrt 2}.$$
But if he did, then Alice and Bob would clearly have classical correlation only: either both ’s (if Charlie got the measurement outcome) or both (if Charlie got). From this it follows that Alice and Bob have only classical correlation regardless of whether Charlie measured or not
As the post-measurement composite state of Alice and Bob's qubits would be either $|0\rangle_A|0\rangle_B$ or $|1\rangle_A|1\rangle_B$, the resulting correlation becomes trivial. If Alice's qubit is measured to be $|0\rangle_A$ or $|1\rangle_A$ then Bob's qubit would be $|0\rangle_B$ or $|1\rangle_B$ respectively and vice-versa.
Moreover, even if say Charlie didn't measure his qubit, if Alice measures her qubit to be $|0\rangle_A$ (or $|1\rangle_A$), Bob's qubit will also be $|0\rangle_B$ (or $|1\rangle_B$). This is again a trivial classical correlation between Alice and Bob. The fact of the matter is that to observe some non-trivial correlations you need to determine the joint statistics of Alice, Bob and Charlie. Pairwise statistics will only show you trivial classical correlations. The magic of entanglement only manifests in the three-body measurement statistics for the GHZ state!
In the GHZ case, we noticed that the effects of entanglement are only observable in the three-qubit statistics and not in the pairwise statistics. Mathematically, it can be shown that in the GHZ state there is no pairwise entanglement at all (from the reduced density matrix $\rho_{AB}$). The "Monogamy of Entanglement" part is a bit of a non-sequitur, indeed. Aaronson states the example of the $|W_3\rangle$ where no two qubits are pairwise "maximally entangled" but are entangled to some degree.
The point is that, in a three-qubit state, it is theoretically impossible for more than one pair of qubits to be maximally entangled. The $|\mathrm{GHZ}_3\rangle$ and $|W_3\rangle$ states are simply examples of this; the interesting fact being that the $|\mathrm{GHZ}_3\rangle$ state as a whole is maximally entangled although no two qubits are pairwise maximally entangled. Thus, it is analogous to Borromean rings which consist of three topological circles which are linked but where removing any one ring leaves the other two unconnected.
