Convergence in SOT and norm boundedness in $C_r^*(S_\infty)$ equivalent to norm convergence Let $S_\infty$ - permutation group of the natural numbers fixing all but a finite number of element. And let $C_r^*(S_\infty)$ - reduced group $C^*$-algebra that acts on $\ell_2(S_\infty)$ in obvious way. 
Let $A_n \in C_r^*(S_\infty)$ is a sequence of operators that bounded by norm (i.e. exist $C>0$ such that $||A_n||<C$ for each $n$) and $\operatorname{lim-SOT} A_n = A_\infty$ (SOT - strong operator topology) is that true, that $\operatorname{lim} A_n = A_\infty$ (in norm topology)?

Definitely no. There is no infinite-dimensional context where you could make that kind of deduction. 
If that property were true, the unit ball of $C_r^*(S_\infty)$ would be SOT-closed, which would make $C_r^*(S_\infty)$ a von Neumann algebra; and this is impossible, since $C_r^*(S_\infty)$ is separable, while any infinite-dimensional von Neumann algebra is not norm-separable. 
Comment 1. The sot-closure of $C_r^*(S_\infty)$ is very very very well-known: it is the hyperfinite II$_1$-factor.
Comment 2. The group $S_\infty$ plays absolutely no role here. For any countable infinite group, the answer will be no.
Comment 3. The assertion fails even in a von Neumann algebra. For instance, consider the von Neumann algebra $B(H)$ with $H$ separable. Fix an orthonormal basis $\{e_n\}\subset H$ and let $E_n$ be the orthogonal projection onto the span of $e_n$ (that is, for each $n$ $E_nx=\langle x,e_n\rangle\,e_n$). Let $A_n=\sum_{k\geq n}E_k$. Then $A_n\to0$ sot, but $\|A_n\|=1$ for all $n$. 
