I'll make a few theoretical remarks which might be useful in computation.
As I said in comments, an Abeliean Frobenius complement is necessarily cyclic.
Given a cyclic group $A$ of order $d$, here's a strategy for determining Frobenius groups with complement $A$ and elementary Abelian kernel $V$ which is a $p$-group for some prime $p$. It is enough to consider the case that $A$ acts irreducibly on $V$ (I will return to this point later).

It is of course necessary that $p$ does not divide $d$. Let $e$ be the smallest positive integer such that $e|d-1$. Then we may construct a Frobenius group with kernel $V$ of order $p^{e}$ and complement isomorphic to $A$ as follows:

The polynomial $\Phi_{d}(x) \in \mathbb{Z}[x]$ reduces (mod $p$) as a product of 
$\frac{\phi(d)}{e}$ irreducible polynomials of degree $e$. Given any one of these irreducible factors, say $m(x)$, we may let $A$ act on $V$ as a linear transformation whose matrix is the companion matrix of $m(x)$. In this way, we obtain $n(d) = \frac{\phi(d)}{e}$ non-isomorphic Frobenius groups with complement isomorphic to $A$ and kernel elementary Abelian of order $p^{e}$.

Then it is easy to check that there are $\frac{n(d)^{2}+n(d)}{2}$ non-isomorphic Frbenius groups with complement isomorphic to $A$ and kernel elementary Abelian of order $p^{2e}$, and similar results hold for elementary Abelian kernels of order $p^{me}$ for any positive integer $m$.

This strategy suffices to construct all non-isomorphic Frobenius groups with cyclic complement and Abelian kernel of squarefree exponent to any specified order.

Dealing with other kernels (eg non-Abelian or Abelian, but not of squarefree exponent) requires more work, as indicated in comments, but for groups of relatively small order this should still be manageable.