I am not sure I understood well your question, but if all weights are positive, I think that you can support weights updates in time within $O(n)$ (better than $O(n^3)$ per update, which I understand is what you asked?).
Consider
- the symmetric matrix $W$ (as in Weight) which stores, for each pair $\{i,j\}$, the weight of this edge;
- the symmetric matrix $T$ (as in Total) which stores, for each pair $\{i,j\}$, the total weight of the minimal path from $i$ to $j$; and
- the symmetric matrix $N$ (as in Next) which stores, for each pair $\{i,j\}$, the first vertex $k$ of the shortest path from $i$ to $j$ (the entire path can then be extracted by checking $\{k,j\}$ recursively). If there are several minimal paths, take the first in lexicographical order if $i<j$ and the last in lexicographical order if $i>j$ to avoid forming cycles.
Initialize $N$ in time within $O(n^3)$ (or better via faster matrix multiplications algorithms). The values of $N$ form a spanning tree of at most $n$ edges and no shortest path contains twice the same edge.
Consider an update decreasing the weight of the edge $\{i,k\}$. For each entry $j$ of the row $i$ in $N$,
if $N\{i,j\}=k'\neq k$ and $W\{i,k'\}+T\{k',j\} > W\{i,k\}+T\{k,j\}$,
- update $N\{i,j\}\leftarrow k$ and $T\{i,j\}\leftarrow W\{i,k\}+T\{k,j\}$;
- update the other entries of $T$ by following the paths involving the edge $\{i,j\}$, using the symmetry of $N$.
Otherwise, do nothing:
- if $N\{i,j\}=k$, given that the weight is decreasing, this particular path is still optimal and does not need to be updated;
- if $N\{i,j\}=k'\neq k$ and $W\{i,k'\}+T\{k',j\} < W\{i,k\}+T\{k,j\}$, the decrease was not important enough to change the minimum path.
Given the spanning tree structure of the solution of the all pair minimum path and the fact that the weights are always decreased, I think that this gives linear time per update. I think you could easily get the same result for directed weighted graphs, at the cost of some extra space and a more complicated structure.