Is it known whether any element of trace 0 in the reduced $C^*$algebra of a nonabelian free group, is a limit of sums of (additive) commutators?

This is true. Perhaps it is known whether this property for a group $G$ is equivalent to $C^*_\lambda G$ having a unique trace? In any case, the same proof of Powers which shows unique trace for free groups, can be adapted to this question. First note that if $F = F(x_1, x_2, \ldots, x_n)$ is the free group on $n$ generators, then by a trick of Bob Powers, in $\mathcal B(\ell^2 F)$ we have $\ \Sigma_{i = 1}^n \lambda(x_i) \ \leq 2 \sqrt{n}$. Indeed, if we denote by $P_i$ the projection in $\mathcal B(\ell^2 F)$ onto the subspace generated by Dirac functions on all words which in reduced form begin with the letter $x_i$, then we have $(1  P_i) \lambda(x_i) (1  P_i) = 0$ and hence $$ \Sigma_{i = 1}^n \lambda(x_i) = \Sigma_{i = 1}^n P_i \lambda(x_i) + ( \Sigma_{i = 1}^n P_i \lambda(x_i) (1  P_i))^*. $$ Since $P_i$ have orthogonal ranges we have that $\ \Sigma_{i = 1}^n P_i \lambda(x_i) \ \leq \sqrt{n}$ and $\ \Sigma_{i = 1}^n P_i \lambda(x_i) (1  P_i) \ \leq \sqrt{n}$. Next note that the limits of sums of commutators forms a subspace and hence it is enough in the reduced group $C^*$algebra to show that the nontrivial group elements can be written as a limit of a sums of commutators. Moreover, in a free nonabelian group, every element has an element which is free from it, and hence by restricting to a subgroup it is enough to show that in the free group on 2 generators $F(a, b)$ we can write $\lambda(a)$ as a limit of sums of commutators. For this we consider an arbitrary $n \in \mathbb N$ and easily verify the formula $$ \frac{1}{n} \Sigma_{i = 0}^{n 1} \lambda( b^{i} a b^i ) = \lambda(a)  \frac{1}{n}\Sigma_{i = 1}^{n  1} [ \lambda(b^i), \lambda(b^{i} a) ]. $$ Since $x_i \mapsto b^{(i  1)} a b^{(i  1)}$ extends to an isomorphism between $F(x_1, \ldots, x_n)$ to the subgroup $\langle a, b^{1} a b, \ldots, b^{(n  1)}a b^{(n  1)} \rangle$ it follows from Powers trick that $$ \ \lambda(a)  \frac{1}{n}\Sigma_{i = 1}^{n  1} [ \lambda(b^i), \lambda(b^{i} a) ] \ \leq 2/ \sqrt{n}. $$ Since $n$ was arbitrary, this finishes the proof. 


More is true: an element of a C*algebra is a norm limit of sums of commutators if and only if it is 0 on any bounded trace. For selfadjoints, this was proven by Cuntz and Pedersen (in their only paper together I think). One reduces from arbitrary elements to selfadjoints by writing $c=a+bi$, with $a$ and $b$ selfadjoint. So if $C^*_r(G)$ has a unique tracial state then any element vanishing on it is a limit of sums of commutators. A couple more facts proven by Cuntz and Pedersen: (1) the commutators can be arranged to form a convergent series, (2) for selfadjoints, the commutators can be chosen of the form $x^* xx x^*$. 

