For each real $k>0$, \begin{equation}E\psi_\infty(|X|/k)=\infty\,P(|X|>k)+P(|X|=k) \\ =\left\{\begin{aligned}\infty\text{ if } P(|X|>k)>0,\\ P(|X|=k)\le1\text{ if } P(|X|>k)=0. \end{aligned}\right. \end{equation} So, indeed, $\|X\|_{\psi_\infty}<\infty$ iff $X$ is essentially bounded. Moreover, $\|X\|_{\psi_\infty}=\text{ess}\,\text{sup}\,|X|$.
Generally, for any non-constant nondecreasing convex function $F\colon[0,\infty)\to[0,\infty]$ such that $F(0)\le1$, the formula \begin{equation}\|X\|_F:=\inf\{t>0\colon EF(|X|/t)\le1\} \end{equation} defines a norm on the linear space, say $L_F$, of random variables (r.v.'s) $X$ on a probability space $\mathcal P$ with $\|X\|_F<\infty$. The proof of this is the same as the one in the case when $F$ is not allowed to take the value $\infty$. (If, in addition, it is assumed that $F(0)<1$, then all bounded r.v.'s on $\mathcal P$ will be in $L_F$.)