A partial solution: If $3n+1\ge 13$, then the board is completely covered by the $4\ (3n+1-6)\times(3n+1-6)$ squares in the corners, so we can assume the missing piece comes from one of those, we can cover that with $L$'s, according to the induction hypothesis (we assume we've shown it's true for smaller boards), and we can cover the rest with $2\times 3$ rectangles (this is easy to see). If $3n+1=10$, the board is covered by the $9\ 4\times 4$ squares which have even distances from their sides to the sides of the board. We can cover any of these $9$ squares with one small square missing according to teh case $3n+1=4$, and for each one of the $9$ squares, the rest can, again, be tiled with $2\times 3$ rectangles. If $3n+1=4$, it's easy to see that a $4\times 4$ minus a $2\times 2$ in a corner is tileable with $L$'s, so we can first place an $L$ near the missing small square so as to complete a $4\times 4$ minus a $2\times 2$ in a corner, and then cover the rest with $L$'s. I guess this leaves $3n+1=7$. The case $3n+1=7$ is not hard at all, but it's very hard to explain it here without some pictures. The idea is to complete the hole created by taking out a small square to a $2\times 2$ square, since there are a lot fewer cases to check. In fact, there are three: we can form a $2\times 2$ square in the corner, or one adjacent to a $2\times 2$ in a corner, or one which contains the center of the $7\times 7$ board (there are other possible positions for a $2\times 2$ square, but we can get one of these, because we also have a certain freedom in choosing the $2\times 2$ square we construct by putting an $L$ near the hole). In these three cases, it's easy to get coverings of the rest of the board.