The intensity depends on the average of the pressure squared so it is that average which must be used. Taking the squared bit out of the log bracket leaves you, as you have noted, with the rms pressure which is not the same as the mean of the magnitude of the pressure. The same is true in ac theory where the rms value is used because the power depends on the current squared.
In the end it boils down to the measurement of energy transfer per second and because that measure is proportional to the pressure squared it is that quantity that is averaged giving greater weight in the averaging process to the larger pressure variations.
To get the loudness you must factor in the sensitivity of the ear (response to different frequencies) to your $L_p$ which is a measure of the intensity of a sound wave relative to some chosen standard intensity. The process of factoring in the sensitivity of the ear is a complex process which is usually present as graphs of the type shown here. This is not of course the response of your ears but a "typical" ear for a person of a certain age.
So if you increase the intensity of a sound that is within the audible range of frequencies then the sound will become louder but that will not be so for frequencies outside that range.
You make several assertions in your "For example" paragraph which simply aren't true. I'm curious as to your sources for these statements, or are they simply a statement you've formed yourself?
If you change only the frequency of a source (constant amplitude), you most definitely change the intensity of the wave, so the level will change. Plus, you are confusing loudness (a psychoacoustic perception) with level (a logarithmic ratio of intensities). To answer your final question: changing frequency with constant amplitude does change the loudness.
What you may seeing in the physics relationships is $$\beta=\log_{10}\left(\frac{I}{I_0}\right)$$ where $\beta$ is the intensity level in units called bels (decibels would be $10\times$ this), $I$ is the intensity of a wave, and $I_0$ is some reference intensity to define a level of $0$ bels.
While you don't explicitly see the frequency in this formula, it's there. It is true that if you have two sources of equal intensity and different frequency, they will have the same level, $\beta$.
Loudness is affected not only by amplitude and frequency, but also by the ear mechanism and the brain. Equal level sounds of differing frequencies have differing loudnesses. You can research Fletcher-Munson curves.
Best Answer
The first thing to understand is that dB is a logarithm of a ratio. More specifically, a logarithm of a ratio of power/power (which is why dB has no units). A Bel is the base-10 logarithm of a power ratio. A decibel slices each Bel into ten parts.
Power is power. It does not matter whether it is sound or electricity or whatever. And dB always refers to a ratio of power.
In most sound references, "dB" means "dB in reference to (as a ratio over) 'X'." Where 'X' is a reference level. I think the reference used is the sound pressure which is the threshold of human hearing. So a sound with 10 times the POWER of that reference would be 10 dB. A sound with POWER 100 times the reference would be 20 dB.
The answer to your question is: It can be. Sound intensity (if expressed as a pressure) and Sound Pressure Level (SPL) are the same. But neither is power. Sound power is proportional to sound pressure squared. Thus, if you increase sound PRESSURE by a factor of 10, you have increased POWER by a factor of 10^2 = 100. These two actions are synonomous, and result in a 20dB increase. Note that I did not explicitly state what increased by 20dB in that last sentence... because it is ALWAYS power.
So, you need to watch your units, and make sure the other guy is watching his. If he doubles his POWER level, that is a 3 dB increase. If he doubles his sound PRESSURE, that is a 6dB increase. If he doubles his sound INTENSITY, you need to nail down whether that is in units of pressure or power.
I hope that helped.