Well, I wasn't so much offering insight, as offering to take a couple measurements on my setup. Oh, and I was offering EE-jargon, too.

But if you think it's insightful, here's more:

Overview:
The sound hardware in the N800 has some stuff which generates a digital output.
This digital output gets run through a DAC (Digital -> Analog Converter). The DAC outputs a voltage corresponding to the numeric input.
This analog signal then goes through some filters, out the headphone jack, and presumably into a load.

So parameters of the system include:
Maximum output sample rate: This is how often the digital output gets updated. Probably 48kHz, but IDK. This is the most important factor in maximum frequency. (The highest output frequency is one-half the sample rate, since you need a positive and a negative peak in each cycle.)
DAC resolution: This is how many levels of voltage the DAC can output.
Probably 16 bits, maybe less. Highly unlikely to be more, though PC sound cards are readily available higher. This is the limiting factor on dynamic range.
Analog filtration stuff:
Since the DAC outputs a constant voltage for a given input, the output has stair-steps. This results in HF noise, at frequencies at and above the sample rate. Those are well above the range of human hearing in this case, but are still probably eliminated with a low-pass filter, which attenuates frequencies above about 24 kHz (sample rate / 2).
There could be more analog filters here; probably volume control is here (it definitely should be), and there may be circuits to compensate for the frequency-dependence of the load.

Finally, the output goes to the headphone jack. Now, if the outputs are directly connected, it'll work fine for headphones. But you could have trouble when connecting it to an amp, as the sound wave is riding on top of a DC level, and the amp may have its inputs referenced to ground. To avoid such trouble (which could cause anything from distortion to frying one of the components), the output is connected to the jack with a capacitor.

Brief discursion into capacitors and inductors:
Resistors have a given resistance value. It's constant for all frequencies of voltage/current you might care to apply. Capacitors and inductors (the ideal ones) have no resistance, but have impedance, which is like resistance. The distinction is that resistance doesn't change the phase; the voltage across it is in phase with the current through it. Impedance shows the same current-voltage relation (high impedance = low current for a given voltage, or high voltage for a given current), but also contains a phase difference between voltage and current. The interesting thing, for our purposes, is that the capacitor has an infinite impedance at DC, and a low impedance at HF. Inductors are the opposite.

So, our capacitor looks like an open circuit for DC -- no DC gets sent through, and if the load has a DC bias, it doesn't send any DC back to fry the N800. For high frequency, it looks like the jack is wired directly.
The only question is, at what frequency does the capacitor "start" conducting?

For the moment, we'll treat your load (earphones, external amp, speakers, whatever) as a resistor. It's not, of course, though the assumption is probably respectable for the amp. The others have strong inductance, but we'll ignore that in the interest of simplicity.

The output power will be one-half the pass-band power (-3dB) when the impedance of the capacitor is equivalent to the resistance of the load. This frequency (called the corner frequency) can be shown to be characterized as 1/(R*C), where R and C are the resistance and capacitance. This formula gives the ouput in radians/second, but we want Hz, so it's 1/(2*pi*R*C) Hz. Above this frequency, we can treat the capacitor as conducting, since the resistance of the load rapidly dominates. Below this, the capacitance dominates, so we can treat the response as falling off at 20 dB/decade. A decade is a power of 10, so if the corner frequency is 100 Hz, a 10 Hz signal would be attenuated by 20 dB, or have one-tenth the voltage, or one-hundredth the power, of a higher-frequency signal with the same amplitude. A graph of the output on a dB scale versus the frequency on a log scale looks like a straight line rising at 45 degrees up to the corner frequency, then level the rest of the way, hence the term corner frequency.

The corner frequency, then, is what we need to find. But it depends on the resistance of the load. So the only thing to do is hook the load (your earphones) up, and measure the voltage output while playing a bunch of sound files containing sine waves with the same amplitude (but a range of frequencies). Look for the point where the output really starts falling off as the frequency drops, and you know the corner frequency. (To best determine it, plot all points on a dB-log plot, giving you the entire frequency response, but if you just want the corner frequency, look for where the output amplitude is drops to 70.7%)
That is easiest with a scope, though you could do it with a good AC voltmeter.

HTH!