Yeah I have read some similar statements... have yet to run across a good explanation.
I suppose that it may depend in part on whether the "single phase mid-tap" is grounded to create a high-leg delta or is used in conjunction with one end of that same secondary winding being grounded to create just one 120 volt supply point.
If the latter case is considered, look at the extra current in that part of the winding from a single phase 120V load:
240V delta will have a total kVA rating of three times 240xAx3 where A is the current in a single delta
winding of the secondary. (Not the current in the phase line, where root 3 would come into play.)
The power from a single-sided 120V load will be 120xA where A is the current in the winding supplying the 120V load, not considering for the moment the parallel current source path through the remainder of the delta. Since that path will have five times the internal resistance, that is a fair approximation.
Notice that for a given load current, A, the power to the 120 volt single phase load will be 1/6 of the power the same winding current could deliver when working to supply a balanced three phase load.
Presto: If you reduce the three phase current by the amount of the 120V load, you end up reducing the available power by six times that amount.
So 5% of the rated kVA going to the 120V load cost you 30% of the available three phase power.
That calculation works for me as a good first approximation, and I would not be surprised if the exact calculation comes out the same.
It all happens because the winding conductor for the tapped winding is no larger than the winding conductors in the rest of the windings.
To the extent that the localized winding current rather than the overall heat dissipation is the limiting factor for the transformer, I think that you have your answer.