The key thing to keep in mind is that the phases of the different signals (including the room reflections) are effectively random. In addition to exact frequency matching, complete destructive interference requires that the the two signals be $180^\circ$ out of phase. Because of the phases of the different signals are uncorrelated with one another, the expected power at the listener's position grows with the number of sources.
Consider the signals in the Fourier domain, the phase of the signal at a particular frequency will be effectively random. As you add more of them together, the amplitude at a particular frequency will undergo a random walk in the complex plane. Thus, the intensity (amplitude squared) at a particular frequency will tend towards the sum of the intensities of the individual components.
If we have an ensemble of sources, producing an complex amplitude at a particular frequency $a_j(\omega) e^{i \phi_j(\omega)}$ Note: I'm not requiring that the sources be pure tones, only that we're just considering one frequency ($\omega$ with associated wave number $k$) at a time, thus I'll drop the explicit $\omega$ dependence. The key point is that the $\phi_i$ is uniformly distributed on $[0,2\pi)$ and independent from source to source.
The complex amplitude at an ear, in an unenclosed space will be:
$A = \sum_j a_j e^{i (k d_j +\phi_j)}$
where the phase includes both the intrinsic phase of the source, and the phase accumulated over the path from the source to the listener.
Your intuition would seem to be vindicated in that if we average over the random phases, we end up with $\langle A \rangle =0$ (I'm using angle brackets for the averaging over the random phases. But this is deceptive because it is zero due to phase symmetry, i.e. the phase of $A$ is uniformly distributed. Whats more interesting is to look at $\langle \lvert A \rvert^2 \rangle = \sum_j a_j^2$ (the cross-terms average out to zero) -- that is the expected power of the net signal grows with the number of sources. One would get a similar result if you examined the expected amplitude $\langle \lvert A \rvert \rangle$ but the math is slightly harder.
Except for special arrangements of the sources w.r.t. the walls of the room, the reflections off the wall also have, effectively, random phases due to the different path lengths from the source-wall-ear (or source-wall-wall-ear and so on), these would appear in the equation as
$$ A= \sum_j a_j e^{i (k d_j + \phi_j)} + \sum_p \sum_i a_j r_{pi} e^{ i( k d_{pj}+\phi_j)} $$
where the additional sum is over the paths, indexed by $p$, with associated reflectivity coefficients $r_{pj}$ and path lengths $d_{pj}$. Note that for typical treble sounds have wavelengths on the order of a meter (or so), and rooms (esp. concert halls) are scaled at 10s of meters, so different paths will not have similar phases as the direct path.
Disclaimer
I can't absolutely, positively rule out the possibility that there might be, for particularly strong singers, some phase-locking between singers (c.f. this video for a mechanical example) and thus the assumption of independent $\phi_j$ might be invalid. Even so, for most musical situations, the path-length phases are large enough to effectively randomize the phases at the location of the listener.
