Does the order in which qubits are measured matter in quantum computing?
No, the order doesn't matter.

Proofs


Algebra. Take an input state $\sum_k c_k |k_0 k_1 k_2 ...\rangle$. Apply the definition of measurement from your textbook to it. Compute the expression for the probabilities and outputs of each case when measuring qubit 0 then qubit 1. Do the same for measuring qubit 1 then qubit 0. Notice that the two expressions are equal. Generalize.
Circuit moves. Measurement is equivalent to a CNOT gate from the target qubit onto an ancilla qubit that you simply don't use for anything else:



So if you think you can change the outcome statistics by reordering measurements, you should think you can do the same without involving measurements at all (until a simultaneous measurement of all qubits at the end of the circuit). All you should need is independent CNOTs. But clearly independent CNOTs can be re-ordered.
Play. For me, the fact that ordering doesn't matter wasn't proven so much as experienced. I dragged gates around in Quirk and noticed that measurement is easily the most boring operation. All measurement does is throw away off-diagonal density matrix elements. Even for the target qubit, measurement doesn't change the computational basis probabilities.