The key properties of the Levi-Civita symbol are its antisymmetry and normalization, $\epsilon_{1\cdots n}=1$.
We wish to capture, with the integers $\{i_1,\ldots,i_n\}$, $i_j\in\{1,\ldots,n\}$, these properties of $\epsilon$.
It is natural to consider products of the form $\prod_{j,k}(i_j-i_k)$, since the product vanishes if any of the $i$s are the same.
In addition, we can't have an even number of factors of the form $(i_j-i_k)$, since we'll have no hope of capturing the antisymmety property of $\epsilon$.
A simple ansatz is
$$\begin{eqnarray*}
e(i_1,\ldots,i_n) &=& c_n\prod_{1\le j<k\le n}(i_k-i_j) \\
&=& c_n \prod_{k=2}^n\prod_{j=1}^{k-1}(i_k-i_j) \\
&=& c_n [(i_2-i_{1})] \\
&& \times [(i_3-i_1)(i_3-i_2)] \\
&& \cdots \\
&& \times [(i_{n-1}-i_{1})\cdots(i_{n-1}-i_{n-2})] \\
&& \times [(i_n-i_{1})\cdots(i_n-i_{n-1})]
\end{eqnarray*}$$
where $c_n$ is some constant that we'll determine shortly.
From the form of the product we can see that permuting adjacent $i$s in the product will simply introduce a factor of $-1$.
(For example, $e(i_2,i_1,\ldots,i_n) = -e(i_1,i_2,\ldots,i_n)$ since we'll pick up one factor of $-1$ from the factor $(i_2-i_1)$.)
This is enough to show that the product has the antisymmetry property of $\epsilon$.
If any of the $i$s are not distinct the product is zero.
All other products can be obtained by permutations of the product
$e(1,\ldots,n)$.
All that remains is to determine $c_n$.
We have
$$\begin{eqnarray*}
e(1,\ldots,n) &=& c_n \prod_{k=2}^n\prod_{j=1}^{k-1}(k-j) \\
&=& c_n \prod_{k=2}^n\prod_{l=1}^{k-1}l
\hspace{5ex} (\textrm{let }m=k-j) \\
&=& c_n \prod_{k=2}^n (k-1)! \\
%&=& c_n \prod_{k=1}^{n-1}k \\
&=& c_n \, \mathrm{sf}(n-1),
\end{eqnarray*}$$
where $\mathrm{sf}(n)=\prod_{k=1}^{n}k!$ is the superfactorial.
(Starting from $n=0$, the sequence of superfactorials is
$1,1,2,12,288,\ldots$.)
Therefore,
$$\begin{eqnarray*}
\epsilon_{i_1\cdots i_n} &=& e(i_1,\ldots,i_n) \\
&=& \frac{1}{\mathrm{sf}(n-1)}
\prod_{1\le j<k\le n}(i_k-i_j).
\hspace{10ex} \textrm{(1)}
\end{eqnarray*}$$
Some results for small $n$ follow.
$n=2$
$$\epsilon_{ij} = j-i, \quad i,j\in\{1,2\}$$
$n=3$
$$\begin{eqnarray*}
\epsilon_{ijk} &=& \frac{1}{2}(j-i)(k-i)(k-j), \quad i,j,k\in\{1,2,3\} \\
&=& \frac{1}{2}(i-j)(j-k)(k-i)
\end{eqnarray*}$$
$n=4$
$$\begin{eqnarray*}
\epsilon_{ijkl} &=& \frac{1}{12}(j-i)(k-i)(k-j)(l-i)(l-j)(l-k),
\quad i,j,k,l\in\{1,2,3,4\}
\end{eqnarray*}$$
Degeneracy
Since $\epsilon_{i_1\cdots i_n}^{2m+1} = \epsilon_{i_1\cdots i_n}$ for $m\in\mathbb{N}$, we immediately find an infinity of other possible representations for $\epsilon$, that is, the product
$$\begin{eqnarray*}
\frac{1}{\mathrm{sf}(n-1)^{2m+1}}
\prod_{1\le j<k\le n}(i_k-i_j)^{2m+1}
\end{eqnarray*}$$
is also a perfectly good representation of $\epsilon$.
The principle of parsimony dictates that representation (1) is preferable.
