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Yes, I think so, too. But I cannot find an example that the norm depends on $n$. I tried high-demension ball and hypercube. I cannot come up with other isotropic high dimensional distributions which are easy to calculate the norm. You give an upper bound, but how to show the bound can be reached?
Oh, it's not this meaning. It should mean that fix $\|X\|_{\psi_2}=K$, $\|\|X\|_2-\sqrt n\|_{\psi_2}$ can still be arbitraily large (that is, it depends on $n$). But when $X_i$ are independent, $\|\|X\|_2-\sqrt n\|_{\psi_2}$ can be bounded by constant.
I am sorry that I didn't give the definition of $\|\cdot\|_{\psi_2}$ of a random vector. It has the same meaning as @Guillaume Aubrun said. I have added it in the question.