Consider an arbitrary $n$-qubit state $\lvert \psi \rangle$. How much do we understand about the probability distribution of the fidelity of $\lvert \psi \rangle$ with a tensor product $\lvert \alpha \rangle = \lvert \alpha\_1 \rangle \lvert \alpha\_2 \rangle \cdots \lvert \alpha\_n \rangle$ of Haar-random pure states $\lvert \alpha\_j \rangle$?

It seems to me that the mean fidelity will be $1/2^n$ (taking fidelity to be measured in units of probability, *i.e.*, $F(\lvert\alpha\rangle,\lvert\psi\rangle) = \lvert \langle \alpha \vert \psi \rangle \rvert^2$). For instance, we can consider the fidelity instead of $\lvert \psi \rangle$, subject to a tensor product of Haar-random single-qubit unitaries, with the state $\lvert 00\cdots0 \rangle$. It seems to me that, up to phases, the Haar-random unitaries on $\lvert \psi \rangle$ will in effect randomise the weights of the components of $\lvert \psi \rangle$ over the standard basis. The expected fidelity with any particular standard basis state would then be the same over all standard basis states, *i.e.* $1/2^n$.

Question.
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What bounds we can describe on the probability, that the fidelity will will differ much from $1/2^n$ (either in absolute or relative error) — without making any assumptions about the amount of entanglement or any other features of $\lvert \psi \rangle$?