Trace of 3 gamma matrices in 3 dimensions I know that for a Lorentzian metric with signature $(-,+,+,..,+)$, in an even number of dimensions the trace of any odd number of gamma matrices is zero.
This can be proven by defining $\gamma_\ast=\gamma^0\gamma^1..\gamma^D$ with $D$ being the number of dimensions, and using the fact that $\{\gamma_\ast,\gamma^M\}=0$ (for even number of dimensions) and $\gamma_\ast^2=1$ (any number of dimensions).
What about the trace of an odd number of gamma matrices in an odd number of dimenions?
 A: I seem to disagree with the answer of G. Smith,
In three dimensions the gamma matrices can be taken as the Paui matrices $\sigma_1,\sigma_2, \sigma_3$ Then  ${\rm tr}\{\sigma_1\sigma_2\sigma_3\}= 2i\ne zero$.  In any odd dimension where $\gamma_{2n+1}=(-i)^{n} \gamma_1\ldots \gamma_{2n}$ has to be included we will have
$$
(-i)^n {\rm tr}\{ \gamma_1,\ldots, \gamma_{2n+1}\}=  2^n.
$$
A: One answer already gave helpful advice but I wanted to point out one subtlety arising from some differences of even and odd Clifford algebras, in particular their irreducible representations. Indeed, let $\gamma_i$ be Clifford generators, $i=1,...,n$, i.e. they satisfy
$$ \gamma_i\gamma_j +\gamma_j\gamma_i = 2\delta_{ij} \ . $$
Then there is a volume element $\eta$ which is simply the product of all Clifford generators:
$$ \eta = \gamma_1... \gamma_d \ . $$
You may check that this element anti-commutes with all Clifford generators for even $d$, but commutes for odd $d$. Therefore, if we consider an irreducible representation of the Clifford algebra, $\eta$ will be représented by a multiple of the identity. 
These were just general comments, I will now tackle your specific problem. You may insist on using, in the odd case, a reducible representation in which case $\eta$ is represented by a distinct element. Then it holds what has been replied before and indeed the trace over an odd number of generators will give zero. However, in 3 dimensions, one often uses the irreducible representation, which is by the Pauli matrices, and you may check that if you multiply them you indeed get something that is proportional to the identity. 
A: As Wikipedia’s “Higher-dimensional gamma matrices” explains,

The proof of the trace identities for gamma matrices is independent of dimension. One therefore only needs to remember the 4D case and then change the overall factor of $4$ to $\text{tr}(I_N)$.

Thus if the trace of an odd number of gammas is zero in an even number of dimensions, it must also be zero in an odd number of dimensions.
Wikipedia’s dimension-independent proof is here.
ADDENDUM: Apparently Wikipedia’s argument is incorrect. See @mikestone’s answer.
