That's not true, it depends on the source impedance and load impedance of each split:

Zloadtotal = Zload0 || Zload1 || . . . Zloadn

where "||" means "in parallel with", or = 1 / (1/ZLoad0 + 1/Zload1 + . . . 1/Zloadn)

compared with an open circuit:

dB loss = 20 * log ( Zloadtotal / (Zsource + Zloadtotal))

or compared with a single load:

dB loss = 20 * log (( Zloadtotal / (Zsource + Zloadtotal)) / (Zload0 / (Zsource + Zload0)))

Such that if you have a 100 ohm source and 1K ohm loads, then compared with open circuit each added load yields:

1 -0.8dB

2 -1.6dB

So the loss from the first split is -0.8dB, subsequent splits vs. open circuit are:

3 -2.3dB

4 -2.9dB

5 -3.5dB

into *each* load (that is, every load will "see" a signal at that level). Thus, the loss gets greater in total, but smaller compared with the previous split.

If the nominal loads are more typical 10K line inputs, the loss is much smaller:

1 -0.09dB

2 -0.17dB

3 -0.26dB

4 -0.34dB

5 -0.42dB

The person giving that advice may have been confusing what happens with speakers and power levels, which actually increase 3dB in total power as you double the count of speakers (of the same impedance), assuming your amp doesn't melt, just like adding more light bulbs in parallel in a room makes the room brighter, except for a tube amp which is more complicated as its source impedance is higher as tube amps are designed to match impedance to maximize power transfer so you have to switch output transformer taps for various loads. Man, that was a long sentence.