Let $A = C^*(T_1,T_2,T_3,... | T_i=T_i^*, ||T_i|| \leqslant 1)$ - universal enveloped $C^*$-algebra of countable family of self-adjoint operators. $A$ have standard Schauder basis, which contains all words generated by $T_i$ (i.e. $T_1 T_3 T_2 T_5$ - is basis vector). Is it true, that it is boundedly complete Schauder basis or monotonically boundedly complete Schauder basis?
Boundedly complete Schauder basis $e_i \in X$ it is such basis, that for every sequence $\alpha_i \in \mathbb{C}$ such the partial sums $V_N = \sum_{i=1}^N \alpha_i e_i$ are bounded in $X$, the sequence $V_N$ are converge in $X$.
Monotonically boundedly complete Schauder basis is the same, but we consider only $\alpha_i$ which converge to zero.
UPD: I found some interesting theorem here (page 2), maybe it will help to answering on that question.
I claim that if $A$ is an infinite-dimensional C*-algebra then it does not have a boundedly complete basis.
Proof. Assume that $A$ has such a basis. A space with a basis is separable, so $A$ is separable. Moreover, $A$ being infinite-dimensional, contains an infinite-dimensional abelian sub-C*-algebra. That abelian subalgebra will contain a sequence of norm-one orthogonal elements and such elements span a Banach-space copy of $c_0$. This copy of $c_0$ will be complemented in $A$ by Sobczyk's theorem. However, a space with a boundedly complete basis is isomorphic to a dual space. $c_0$ is not complemented in any isomorphic copy of a dual space, because it would be complemented in its bidual (see the diagram on p. 22 here) but by the Phillips–Sobczyk theorem (or by the Grothendieck property of $\ell_\infty$), no copy of $c_0$ is complemented in $\ell_\infty$.
The above argument shows that there is no infinite-dimensional, separable C*-algebra isomorphic to a dual space as a Banach space. In particular, there is no separable, infinite dimensional von Neumann algebra.