tl;dr In general no, but for bound systems the RMS momentum is within a small factor of the momentum width
Let's start with the title question:
Is the momentum of a microscopic particle always equal to or less than the error of momentum
Certainly not. For example, when you set up a beam of electrons in a CRT their momentum far exceed the uncertainty on their momentum.
But that is not the case your teacher is asking about. He has posed you a problem involving bound (not free) electrons.
That tells us several things, the first of which is that their average momentum relative the nucleus is zero, because otherwise over time they would wander away (contrary to the statement that they are bound). It also tells us that their average position is zero (in the nucleus!), a detail that the instructor has probably neglected to mention. That doesn't actually mean they have any significant probability of being found in the nucleus any more than the average momentum being zero requires that they have much probability of being stopped; it just means that for every positive contribution to the average there is a matching negative contribution.
Once you know that (a) the system has on average zero momentum and (b) the system exhibits a finite spread of momenta, it follows that the average magnitude of momentum (say the root-mean-square momentum) will be only a small factor away from the spread.
Which is the point of this exercise, and the underlying truth that makes this kind of hand-wavy quasi-quantitative argument good for order-of-magnitude and back-of-an-envelope calculations.
Note that any individual measurement of the magnitude momentum might gives values considerably larger than or smaller than width of the momentum distribution. It is only the average magnitude that is constrained.
To actually understand why this works that way you need to know a little bit of the math underlying the uncertainty principle. It is worth saying that his math is not unique to quantum mechanics, it was first noticed in optics, so the weirdness here isn't intrinsically quantum mechanical, but belongs to all theories of waves (optics, acoustics, ...) or that respect a wavelike governing equation (quantum mechanics).
The position and momentum representations are related to one another by a Fourier transform, which means that every value in position that is non-zero contributes (in a positive or negative way) across the whole momentum spectrum and vice versa. The next part of the spacial spectrum can, in principle, cancel out parts of the momentum representation, but this will be done periodically across the whole of momentum space. To get a square-intergrable peak in momentum you need a continuous distribution in space and vice versa. Typically, we consider a square-intergrable distribution in both space; then both peaks are infinite in breadth, but asymptotically approach zero faster than $1/x^2$.
This Fourier relationship also means that measuring the momentum distribution of the components of a system (relatively easy to do in particle physics) lets you deduce the position distribution (hard to measure directly). We used this tool in my dissertation work.