Yes, it's conceivable that with some output stages that are designed inexpensively and not very conservatively, long term reliability could be compromised. Which is why I suggested that the op "first check with the manufacturer of the amp, to make sure that connecting the two outputs together will not cause damage or compromise long-term reliability."
But with typical quality designs (and many cheap designs as well), I would expect that there would be no problem. First, consider that the majority of the energy is typically in the mono component of the signal, and in driving that component the two output stages would not be fighting each other, so to speak, they would be helping each other. In other words, they would both be trying to put out the same voltage, so they would not be loading each other or drawing current from each other.
But let's take a worst case, and say that each channel is trying to put out a large 2-volt signal that is 180 degrees out of phase with the signal from the other channel. And let's say that the output impedance of each channel is 100 ohms. That would result in each output sourcing or sinking a current of 4/200 = 20ma (20 milliamperes). I would expect that to be well within the limits of what a typical quality output driver can handle on a continuous, long-term basis.
With respect to the more subtle sonic effects that may occur due to the load being more difficult, they shouldn't particularly matter. First, because it is only the bass frequencies that we are concerned with, and just the mono component of the bass. Second, if the output impedance of the preamp tends to vary significantly within the bass region (due, perhaps, to a coupling capacitor at the output), that wouldn't matter either because the variation would presumably be the same in the other channel, so the voltage divider ratio between each driver device and the summed output would always be 50-50.
As far as driving the cable is concerned, that would actually be improved. The source impedance driving the cable would be one-half of the output impedance of each channel, per Thevenin's Theorem