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 $m_p$.
You measure mass by observing it's acceleration response to force (i.e by applying Newton's second law).
Now, because it is impractical to accurately measure straight-line accelerations over a wide range, we actually use periodic motions and measure frequency.
- Mass-on-a-spring harmonic oscillator. $\omega = \sqrt{\frac{k}{m}}$ with known spring constant.
- Measure the centripetal force on a centrifuge. $F = m \frac{v^2}{r} = m \omega^2 r$, is the naive approach, but on the surface of the planet you have to be a little more clever (adding the centripetal force to the existing weight). Here you would put a scale between the test mass and the centrifuge to get $F$.
An alternative is to measure both the weight and the local value of $g$, which can be done with a small-angle pendulum ($\omega = \sqrt{g/\ell}$).
Best Answer
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.