How to find long trails in a multidigraph
Here's a suggestion for a simple greedy heuristic to apply to the big SCC. It's based on the observation that a cycle of edges can be traversed a number of times equal to the minimum multiplicity of any edge in the cycle, so looking for cycles in which all edges have large multiplicity may be a good strategy:

Order edges in the big SCC in decreasing order by multiplicity. Call this list of edges $X$. (Note: Each edge appears in $X$ just once, regardless of its multiplicity.)
Set $Y$ to be an empty list of edges.
If no edges remain in $X$, go to 7.
Choose the edge $e$ with highest multiplicity remaining in $X$. Remove it from $X$ and add it to $Y$.
Test whether $Y$ now contains a cycle (since this cycle must contain $e$, this check can be done quickly using a union/find data structure): If so, let $Z \subseteq Y$ be the edges in this cycle. Add the pair $(Z, m)$ to the set $C$, where $m$ is the minimum multiplicity of any edge in $Z$, clear out $Y$ and start over (recurse) with $X\setminus Z$.
Otherwise, go to 3.
At this point, we have a set of edge-disjoint cycles in $C$, as well as the maximum number of times each can be completely traversed. Note that if $(Z, m) \in C$, then the cycle $Z$ can be entered at any of its vertices, traversed $m$ times, and then exited at the same vertex; or entered at any vertex, traversed $m-1$ times, then (possibly after a further partial traversal) exited from any vertex, giving us some flexibility.
Find the vertex $v_1$ that is shared by the largest number of cycles in $C$; let these cycles be $C_1 \subseteq C$. If $C_1 = C$ then stop here: We can start at $v_1$ and traverse each cycle the full number of times in any order.
Otherwise, find the vertex $v_2$ that is shared by the largest number of cycles in $C\setminus C_1$ and is adjacent to $v_1$. Repeat this for $v_3$, etc., until $C$ is exhausted.
Starting at $v_1$, for each $i$ in increasing order, and for each cycle $C_{ij} \in C_i$ (in no particular order), add either $m_{ij}$ or $m_{ij}-1$ traversals of $C_{ij}$ to the the trail, whichever is larger and still allows us to stop at $v_{i+1}$ in preparation to begin traversing the following set of cycles.

Since the procedure is fairly quick, it may be worthwhile to rerun it many times, choosing a different order of edges each time, and pick the best overall.