Let $|\Phi\rangle$ be a normalized vector in $\mathbb{C}^d$ and let $|\psi\rangle$ be a random stabilizer state. I am trying to compute the quantity
$$\mathsf{Pr}\big[|\langle \Phi|\psi \rangle|^2 \geq \epsilon \big].$$
Note that if $|\psi\rangle$ is Haar random, then, by equation $2$ of this paper,
$$\mathsf{Pr}\big[|\langle \Phi|\psi \rangle|^2 \geq \epsilon \big] \leq \mathsf{exp}(-(2d-1) \epsilon).$$
Does a similar concentration bound hold for random stabilizer states too?
No such bound holds for general $|\Phi\rangle$. The set $\mathcal{S}_n$ of $n$-qubit stabilizer states is finite, so $m=\min_{|\psi\rangle\in\mathcal{S}_n} |\langle\Phi|\psi\rangle|^2$ is well-defined and if $|\Phi\rangle$ is not a stabilizer state then $m>0$. But then for any $\epsilon\in[0,m]$ we have $\mathrm{Pr}\left[|\langle\Phi|\psi\rangle|^2\ge\epsilon\right]=1$ which rules out any general bound $\mathrm{Pr}\left[|\langle\Phi|\psi\rangle|^2\ge\epsilon\right]\le f(\epsilon)$ with $f(\epsilon)<1$ for $\epsilon>0$. In particular, no general bound of the form $\mathrm{Pr}\left[|\langle\Phi|\psi\rangle|^2\ge\epsilon\right]\le\exp(-a\epsilon)$ with $a>0$ is possible.