Let
\begin{equation}
\mu_X=\tfrac12\,\mu_{aZ}+\tfrac12\,\mu_{bZ},
\end{equation}
where $\mu_U$ denotes the probability distribution of a random vector $U$, $Z\sim N(0,I_n)$,
and $a,b$ are constants such that
\begin{equation}
0<a<1<b\quad\text{and}\quad \tfrac12\,a^2+\tfrac12\,b^2=1.
\end{equation}
Then $EXX^T=I_n$. Also, for any unit vector $u$ and real $s>0$
\begin{equation}
E\exp\{\left<X,u\right>^2/s^2\}=\frac1{2\sqrt{1-2a^2/s^2}}+\frac1{2\sqrt{1-2b^2/s^2}}<2
\end{equation}
if $s$ is large enough (depending only on $a,b$),
so that, by the definition of $\|\cdot\|_{\psi_2}\|$, we have $\|\left<X,u\right>\|_{\psi_2}\le s$. For instance, here we can take $a=1/5,b=7/5,s=3$.
On the other hand, for
\begin{equation}
t:=(b-1)\sqrt{n}/2,
\end{equation}
\begin{multline}
2\,Ee^{(\|X\|-\sqrt n)^2/t^2}>Ee^{(\|bZ\|-\sqrt n)^2/t^2}
>Ee^{(\|bZ\|-\sqrt n)^2/t^2}1_{\|Z\|^2>n} \\
>e^{(b\sqrt n-\sqrt n)^2/t^2}\,P(\|Z\|^2>n)=e^4\,P(\|Z\|^2>n)\to e^4/2>4,
\end{multline}
because, by the central limit theorem, $P(\|Z\|^2>n)\to1/2$.
So, for all large enough $n$,
\begin{equation}
\|\|X\|-\sqrt n\|_{\psi_2}\ge t=(b-1)\sqrt{n}/2\to\infty,
\end{equation}
as desired.
