A polynomial $f\in K[\underline Y]$ is in the kernel of your map iff $f$ is zero in the quotient $$\frac{k[X,Y]}{\bigl((X_i-X_j)Y_{i,j}-1:1\leq i<j\leq n\bigr)}.$$In other words, your kernel is the intersection of the ideal in the denominator with the ring $k[Y]$, $$\ker\varepsilon=k[Y]\cap\bigl((X_i-X_j)Y_{i,j}-1:1\leq i<j\leq n\bigr).$$This intersection is generated by the elements of a Groebner basis which only contain $Y$s, assuming you are using a monomial order which eliminates the $X$s; this is explained in the book by Cox, Little and O'Shea, for example.
Doing small examples shows that
$(\star)$ the intersection is generated by all polynomials of the form $$Y_{i,j} Y_{i,k}+ Y_{j,k}Y_{j,i}+Y_{k,i}Y_{k,j}$$ with $i$, $j$ and $k$ distinct. (I am identifying $Y_{i,j}$ with $-Y_{j,i}$ here when $i\neq j$)
Ordering the variables as in $$X_1,X_2,X_3,X_4,Y_{1,2},Y_{1,3},Y_{1,4},Y_{2,3},Y_{2,4},Y_{3,4}$$ for $n=4$ we find the Groebner basis $$\begin{array}{l} Y_{2,3} Y_{2,4}+Y_{3,4} Y_{2,4}-Y_{2,3} Y_{3,4} \\\\ Y_{1,3} Y_{1,4}+Y_{3,4} Y_{1,4}-Y_{1,3} Y_{3,4} \\\\ Y_{1,2} Y_{1,4}+Y_{2,4} Y_{1,4}-Y_{1,2} Y_{2,4} \\\\ Y_{1,2} Y_{1,3}+Y_{2,3} Y_{1,3}-Y_{1,2} Y_{2,3} \\\\ X_3 Y_{3,4}-X_4 Y_{3,4}-1 \\\\ X_2 Y_{2,4}-X_4 Y_{2,4}-1 \\\\ X_2 Y_{2,3}-X_3 Y_{2,3}-1 \\\\ X_1 Y_{1,4}-X_4 Y_{1,4}-1 \\\\ X_1 Y_{1,3}-X_3 Y_{1,3}-1 \\\\ X_1 Y_{1,2}-X_2 Y_{1,2}-1 \end{array}$$ The same pattern is seen for all $n$. It is very easy to see that all these polynomials are in $((X_i-X_j)Y_{i,j}-1:1\leq i<j\leq n)$, and it should not be difficult to show that they are a Groebner basis in general. I expect checking that the above claim $(\star)$ can actually be proved without much pain.