This answer is really just a convoluted edition of Fedor's answer.  By Cauchy's theorem,
$$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi
   \frac{(e^{i\theta_1-i\theta_2})^{2a+1}(e^{i\theta_2-i\theta_3})^{2b+1}(e^{i\theta_3-i\theta_1})^{2c+1}}{e^{i(2a+2b+2c)}}
\,d\theta_1\,d\theta_2\,d\theta_3.$$
Advancing each variable by $\pi$ changes the sign of the integrand but not the value of the integral, so its value must be 0.