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According to Wikipedia, the Honeywell computer has quantum volume ("QV") equal to 1024 and has 10 qubits. QV should not be more than $n^2$ squared, i.e. 100.
Is this any math error? Are the publicly given QV useless?
So it appears they are using the new (depth) definition, which to me seems to allow simulation and other rooms to inflate performance.
It is not an error. Under the subsection IBM's modified definition, they define the quantum volume as:
$\log_2 V_Q = \arg\max_{n \le N} \{\min [n, d(n)]\}$
That is, if you take the minimum of the number of qubits and the circuit depth over those qubits (i.e. $10$, because the circuit may well be deeper than the number of qubits, so the number of qubits is the limiting factor on Honeywell's device), that is equal to the $\log_2$ of the quantum volume $V_Q$, so $\log_2 1024 = 10$ or, alternatively, $2^{10} = 1024$.
As far as the utility of publicly given quantum volumes, in my mind they're not useless, rather they just require interpretation and a clear statement on which exact formula is being used. Other approaches to quantum volume exist; I really like this one paper that discusses volumetric benchmarks as a generic framework for understanding the capabilities of quantum hardware to implement specific circuit classes. One of the crucial ideas is that you can have circuits with many different shapes (among other properties) and, in practice, application circuits are pretty much never perfectly square or random.
