When calculating the future orbit of an artificial (Earth) satellite, is the moon's gravity significant or insignificant?
It's a great question!
Of course everyone's definition of "significant" will be different.
Gravity is a long range force, it never goes to zero, and decreases only as $1/r^2$. Nothing blocks it either, so basically everything pulls on everything. The only limit is the speed of light. If something is one million years old but it's 1.1 million light years away, you won't feel its gravity for another 100,000 years.
Every solar system body pulls on every other solar system body. Many asteroids have had their masses determined by their gravitational effect on other asteroids that aren't particularly close for example.
For short term predictions of Earth satellites using SGP4 there is an explicit correction term for the gravitational effects of the Moon and Sun. For satellites with higher orbits meaning closer to the Moon and farther from the Earth so that the Earth's effect is weaker, SGP4 uses what is called its "deep space" correction. This is discussed in more detail in the unanswered question How do “Deep space” corrections in SGP4 account for the Sun's and Moon's gravity?
If you are doing a careful calculation (more than just SGP4) then you will always calculate the effects of the Moon's gravity, and you will say that it is always significant.
Artificial Earth satellites with particularly high orbits can be found in this answer and this answer (they go out past 200,000 km!) and a discussion of an artificial Earth satellite that is carefully designed around the Moon's gravitational attraction/perturbations can be found in this answer to TESS orbit and moon resonance.
When they are making their calculation, is the moon so far away and the satellite has so little mass that the moon's influence on the path can be completely ignored?
No, never! And there's a second point, the mass of the artificial satellite doesn't matter. What we care about is the change in the orbit's velocity which is expressed as an acceleration. Start with $F=ma$ and flip it around to get $a_2 = F/m_2$ where the subscript 2 is for the artificial satellite.
Yes, $F=m_1 m_2/r^2$ but $a_2 = m_1 m_2/r^2/m_2 = m_1/r^2$ where $m_1$ is the mass of the Earth or Moon. If the satellite is 10x heavier, then the force from the Moon is 10x larger but that's cancelled by the 10x larger mass, so the acceleration is the same.
The only (at least almost) time the mass of the artificial satellite becomes important is when it is so massive that it alters the orbit of the Moon. The only other time is when there is a non-gravitational force like drag from the top part of the atmosphere, or solar wind or photon pressure, because those forces do not depend on mass explicitly so there's no cancellation.