You ask, "How do we describe mass to the aliens, who don't know about our (g)?" This is an example of a class of questions referred to by Martin Gardner as "Ozma problems." The classic Ozma problem is how we describe to aliens the distinction between right and left; the answer is that we do it by describing the weak nuclear force.
Your statement of your Ozma problem seems a little ambiguous to me. Essentially you're asking how we describe to the aliens a unit of gravitational mass. (You don't say so explicitly, but it seems clear from context that you don't mean inertial mass.) Futhermore, there is a distinction bewteen active gravitational mass (the ability to create spacetime curvature) and passive gravitational mass (what we measure with a balance). Not only that, but your question could be interpreted as asking whether we can compare with the aliens and see whether the value of the gravitational constant $G$ is the same in their region of spacetime as it is in ours.
We can easily establish 1 g as a unit of inertial mass. For example, we can say that it's the inertia of a certain number of carbon-12 atoms.
The equivalence principle holds for us, so presumably it holds in experiments done by the aliens as well. This establishes that our 1 g unit of inertial mass can also be used as a unit for the passive gravitational mass of test particles.
You didn't ask about active gravitational mass, but the equivalence of active and passive gravitational mass is required by conservation of momentum, and has also been verified empirically in Kreuzer 1968. Cf. Will 1976 and Bartlett 1986.
The other issue is whether $G$ is the same for the aliens as for us. Duff 2002 has an explanation of the fact that it is impossible to test whether unitful constants vary between one region of spacetime and another. However, there are various unitless constants that involve $G$, such as the ratio of the mass of the electron to the Planck mass.
A more fundamental difficulty in the fundamental definition of mass is that general relativity doesn't seem to offer any way of defining a conserved, global, scalar measure of mass-energy. See, e.g., MTW, p. 457
Bartlett, Phys. Rev. Lett. 57 (1986) 21.
Duff, 2002, "Comment on time-variation of fundamental constants," http://arxiv.org/abs/hep-th/0208093
Kreuzer, Phys. Rev. 169 (1968) 1007
MTW: Misner, Thorne, and Wheeler, Gravitation, 1973.
Will, “Active mass in relativistic gravity: Theoretical interpretation of
the Kreuzer experiment,” Ap. J. 204 (1976) 234, available online at adsabs.
harvard.edu.
This is a description of the experiment Cavendish performed at the end of the 18th Century to measure the density of the Earth:
Cavendish put two lead balls on either end of a long bar. He hung the bar at its center from a long twisted wire with known torque. Then, he placed two really massive objects at exactly identical fixed distances from the center of the torsion bar, in the plane of the torsion bar and at right angles to the bar at rest. The balls were attracted and started the wire twisting, but their inertia caused them to overshoot the equilibrium position of the wire. The bar wound up oscillating, and Cavendish measured the rate of oscillation to determine the torsion coefficient of the wire.
With this, he was able to determine the force attracting the balls to each other, which he used to set up a proportion to derive the density of the Earth. Here is a description of the experiment: http://large.stanford.edu/courses/2007/ph210/chang1/, as well as a derivation of the gravitational constant, Big G, that you can perform: http://www.school-for-champions.com/science/gravitation_cavendish_experiment.htm#.VUFS80uiKlI.
One can use the density derived by Cavendish, and the diameter of the Earth (which has been known since Eratosthenes in ancient Greece) to compute the mass of the Earth.
To find the mass of the Earth using the modern form of Newton's Law of Gravitation, you may enploy Little g, the Earth's gravitational acceleration, which is determined by dropping an object, any object, and measuring its acceleration toward the Earth. You do not have to know the mass of the Earth to measure an object's acceleration toward the Earth. Then, you plug the acceleration (9.81 m/sec^2), and the mass of the dropped object into Newton's definition of Force (F=ma), to find the force (F) that the Earth exerts (gravitational acceleration) at the height from which you dropped the object.
Now you know everything in the equation F = g * (m1*m2)/r^2, except for m2, the mass of the Earth. Solve for m2!
Although Newton did not know the magnitude of the gravitational constant (Big G), the form of his equation, which sets the force of gravity inversely proportional to the square of the distance between objects, was rapidly accepted by scientists because it agrees with the motions of the planets as measured by Keppler.
Best Answer
The hardest part is to get started by measuring one particle's mass. Once you have that, you can get the others more easily because ratios of masses are easier to measure than absolute ones. You're quoting values that are good to 8 sig figs, and I'm sure the experiments needed in order to get that kind of precision are extremely complex. I'm going to describe how to measure them with much less heroic precision, using the kind of equipment you could find in a high school with a well equipped physics and chem lab. To this kind of precision, the masses of the neutron and proton are the same.
You say you know how to measure the electron's mass, but just to be concrete, let's say that we'll do that by measuring the wavelengths in the emission spectrum of hydrogen and solving the equations in the Bohr model to get the electron's mass.
Next, we can measure the charge to mass ratio of the electron $-e/m_e$ by accelerating electrons through a known potential and then measuring their deflection in a magnetic field.
Now measure the charge-to-mass ratio of some element such as sodium by doing electrolysis and finding the ratio of how much charge flowed around the circuit to how much mass was deposited on the electrode. Since we know how many neutrons and protons there are in sodium, we can infer the charge-to-mass ratios $e/m_p$ and $e/m_n$ (which, to our precision, are equal).
Finally, since we know $m_e$, $-e/m_e$, $e/m_p$, and $e/m_n$, it's straightforward to find $m_n$ and $m_p$.