Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?
Yes, every connected cubic graph is 3-almost-Hamiltonian. Replace each edge by two parallel edges then follow an Eulerian circuit.
In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.