As $I_0^\prime = I_1$, we have $ I_0\left(\frac{x}{n}\right)^{n-1}I_1\left(\frac{x}{n}\right) = \frac{d}{dx} \left(I_0(\frac{x}{n})^n\right)$ , so
$$
\psi(n) = - n + n\int_0^\infty I_0\left(\frac{x}{n}\right)^n\exp(-x)dx,\\
= -n +n^2 \int_0^\infty (I_0(y) \exp(-y))^n dy
$$

and as $0\leq \exp(-x)I_0(x)\leq1$, expanding $(n+1)^2$,
$$
\psi(n+1)-\psi(n) = - 1 +
(n+1)^2 \int_0^\infty (I_0(y) \exp(-y))^{n+1} dy \\
-(n)^2 \int_0^\infty (I_0(y) \exp(-y))^{n} dy \\
\leq (2n+1)\int_0^\infty (I_0(y) \exp(-y))^{n+1} dy-1.
$$
Now let us use a rough upper bound for $I_0(x) \exp(-x)$, e.g., $$
I_0(x) \exp(-x) \leq \frac{4}{3\sqrt{2\pi x+1}}.
$$
Then
$$
\frac{4}{3}(2n+1)\int_0^\infty(2\pi x+1)^{-(n+1)/2}=\frac{4}{3\pi}\frac{2n+1}{n-1}
$$
and provided $n\geq 10$,
$$
\psi(n+1)-\psi(n) \leq 0
$$