Getting ALL negative weight cycles of a graph using Bellman-Ford
You can use a brute force over all sequences of vertices, check if the permutation represents a negative cycle, and if so- print the cycle. This is not an efficient algorithm (runs at $O(n^{n+1})$), but a simple one 

Suppose all vertices are denoted by $1,2,\ldots,n$. We denote by $C[i]$ the $i^{th}$ node in the cycle, for $i=1,2,\ldots,k$, where $3 \leq k\leq n-1$ is the size of the cycle. 

A pseudo code of the algorithm goes as follows:


Let $k\gets 3$
If $k==n$, then terminate the algorithm.
Let $C[i]\gets 1$ for $i=1,2,\ldots,k$ 
Check if $C=(C[1],C[2],\ldots,C[k],C[1])$ is a negative cycle in the graph. If so- print $C$
If $C[1]=C[2]=\ldots=C[k]=n$, then $k \gets k+1$ and go to step 2.
Otherwise, select $C[j]=\min_i C[i]$ (if there are multiple $j$-s with the minimal value, select the one with the minimal index), assign  $C[j]\gets C[j]+1$ and go to step 4.