What is the escape velocity of a neutron particle (not neutron star) I'm not sure if this question makes sense (if not maybe you can explain why)
But if the neutron has mass and have a size, then it should have a escape velocity in the "surface" right?
I know the gravity force generated by a neutron is really low, since it is also really small in size, what happens with that force when we are really close? is not almost infinite?
 A: 
I know the gravity force generated by a neutron is really low

To define the force, you need to define a second object, with some mass, that is being acted on. That can't be a second neutron, because then there would be an attraction due to the strong nuclear force that would be much greater than the attraction due to gravity. You would want to talk about a particle such as an electron or some other lepton that doesn't participate in the strong force.

what happens with that force when we are really close? is not almost infinite?

The neutron is similar to an object like the earth, in that its mass is distributed over some volume. Therefore you can't get to zero distance from all its mass. The radius of a neutron is roughly 0.8 fm ($10^{-15}$ m). (It's fuzzy, but the number is well defined to within about 20%, if you take some criterion like where the density falls off to half of the value at the center.) Using this radius, the escape velocity is $\sqrt{2Gm/r}=1.7\times10^{-11}$ m/s. The extreme smallness of this velocity confirms that gravity is too weak to matter at the atomic scale.
In reality, if you take a particle such as an electron and try to put it right at the surface of the nucleus, constraining its position to within less than ~1 fm, then by the Heisenberg uncertainty principle, it will be moving many orders of magnitude faster than escape velocity. To get this zero-point velocity to be as small as the escape velocity, you would need a very massive particle -- much more massive than any subatomic particle we know of.
A: I'm sure a full quantum mechanical explanation exists that takes complexities at these size scales into account. But using the classical formula for escape velocity ($v_{esc}=\sqrt{2GM/R}$, derived by determining the total amount of work to move a massive point particle from the surface of a massive object to an infinite distance), an estimate of the escape speed using the neutron mass ($1.67\times 10^{-27}$ kg), and radius ($\sim 1.5$ fm) comes to about $1\times 10^{-11}$ m/s.
