The answer is NO. Rordam and Robert MR3072284 have found
a sequence $(A_n)_n$ of simple unital infinite dimensional C*-algebras
such that $\prod A_n$ has a nonzero character.
(Thanks are due to Yasuhiko Sato for informing me of this.)
Thus the following is true: For every $m$ and $C>1$, there is $n=n(m,C)$
such that $1=\sum_{k=1}^m a_k[b_k,c_k]d_k$ in $A_n$ implies
$\sum_{k=1}^m\|a_k\| \|b_k\| \|c_k\| \|d_k\| > C$.
Now consider the $c_0$-sum $A:=\bigoplus_m A_{n(m,m^2)}$.
Then $(m^{-1})_m \in A$ cannot be expressed as
$\sum_{k=1}^l a_k[b_k,c_k]d_k$ in $A$, because it would imply
$\sum_{k=1}^l \|a_k(m)\| \|b_k(m)\| \|c_k(m)\| \|d_k(m)\| \geq m$ for every $m\geq l$.

On the the hand, it is easy to show that there are $m$ and $C>1$
that satisfies the following:
For every von Neumann algebra without nonzero abelian direct summand,
one has $1=\sum_{k=1}^m a_k[b_k,c_k]d_k$ for some
$a_k,b_k,c_k,d_k$ with $\sum_{k=1}^m\|a_k\| \|b_k\| \|c_k\| \|d_k\| < C$.
By the Hahn--Banach separation theorem, this implies the following:
For every $A$ without nonzero characters and every $x\in A$,
there are infinite sequences $a_k,b_k,c_k,d_k$ such that
$\sum_{k=1}^\infty\|a_k\| \|b_k\| \|c_k\| \|d_k\| \le C\|x\|$
and $x = \sum_{k=1}^\infty a_k[b_k,c_k]d_k$.