Is it possible to construct a Loeb measure for $\{\epsilon: \epsilon\in[0,1], \epsilon \text{ infinitesimal}\}$? Fix an infinite number $\omega$, and a finite number $n$. Let
$$
\Omega =  \left\{\frac{k}{\omega}: k=1,2,\ldots,n\right\},
$$
then $\Omega$ is a subset of the infinitesimal in $[0,1]$.
Is it possible to construct the Loeb measure associated to
$$
X = (\Omega,\sigma(\Omega),\text{ counting measure}),
$$
where $\sigma(\Omega)=2^\Omega$ is the power set of $\Omega$.
That is 
$$
\text{$X$ is a uniform random variable taking value in $\Omega$}.
$$
From this question, I should look for a first order description of $\sigma(\Omega)$.
But I don't see such description.

The answer to the question in the title is no. A Loeb measure is an extension of the standard part of an internal measure defined on an internal measure space. This means that you can't construct any sort of Loeb measure on the set of positive infinitesimals because that set is external.
The answer to the question in the body is yes. Since $n \in \mathbb{N}$, the set $\Omega = \{1/\omega, 2/\omega, \ldots, n/\omega\}$ is finite. An explicit bijection with a finite set is $k\mapsto k/\omega$ for $k = 1,\ldots, n$. Since $2^\Omega$ is already a $\sigma$-algebra (since it's a finite algebra) and the counting measure already takes values in $\mathbb{R}$, the Loeb construction leaves the probability space $(\Omega, 2^\Omega, \text{counting measure})$ unchanged. 
