If $0<s\le 1$, then $D(T^s)\supseteq D(T)$ for any self-adjoint $T$, as is obvious from the description of the domain that you quote at the end of your post.
If $s=n+t$ with $n\in\mathbb N$ and $0<t\le 1$, then $T^s=T^tT^n$. In your case, $T^n=\overline{\Delta}^n$ maps $C_0^{\infty}$ back to itself, and, as just observed, $C_0^{\infty}\subseteq D(T^t)$. Hence $C_0^{\infty}\subseteq D(T^s)$.