It is NP-hard. It is as hard as the standard Hamiltonian path problem (testing the existence of a Hamiltonian path in an arbitrary graph).
Here is the reduction. Suppose you have an arbitrary graph $G$, with no edge weights/lengths, and we want to know whether it has a Hamiltonian path or not. Treat every edge of $G$ as having length 1. Then, add all missing edges, but with edge length of some huge number (larger than $|V|$, the number of vertices in $G$), so now you have a complete graph. Suppose you had an efficient way to find the shortest Hamiltonian path in any complete graph. Then you could find the shortest Hamiltonian path in this complete graph. If its total length is less than $|V|$, then it is a Hamiltonian path for $G$ (it obviously can't include any of the huge-length edges); if its total length is $\ge |V|$, then $G$ has no Hamiltonian path.
Your problem is almost the same as the Travelling Salesman Problem, but for a path instead of a cycle.