In my answer to the same question over on math.stackexchange, I discussed general m by n boards. Here I will just give a winning strategy for Player 2 on the ordinary 8 by 8 chessboard. It's based on the following pairing of squares:
a1c2, b1d2, c1a2, d1b2, e1g2, f1h2, g1e2, h1f2,
a3c4, b3d4, c3a4, d3b4, e3g4, f3h4, g3e4, h3f4,
a5c6, b5d6, c5a6, d5b6, e5g6, f5h6, g5e6, h5f6,
a7c8, b7d8, c7a8, d7b8, e7g8, f7h8, g7e8, h7f8.
That is, square a1 is paired with square c2, b1 is paired with d2, and so on. Note that the 64 squares are partitioned into nonoverlapping pairs, and each set of paired squares are a knight's move apart. The winning strategy for Player 2 is: WHEREVER PLAYER 1 PUTS THE KNIGHT, MOVE IT TO THE OTHER SQUARE IN THE SAME PAIR. For instance, if Player 1 starts the game by dropping the knight on e8, Player 2 replies by playing Ng7. If Player 1 now plays Ne6, Player 2 plays Ng5, and so on. Here is a sample game, with Player 1 making random moves, and Player 2 following the winning strategy described above:
- Ne8 Ng7 2. Ne6 Ng5 3. Nf3 Nh4 4. Nf5 Nh6 5. Nf7 Nh8 6. Ng6 Ne5 7. Nc6 Na5 8. Nb3 Nd4 9. Ne2 Ng1 10. Nh3 Nf4 11. Nh5 Nf6 12. Ng4 Ne3 13. Nc5 Na3 14. Nb1 Nd2 15. Nf1 Nh2 and Player 2 wins.
P.S. Of course the strategy still works if White is allowed to move the knight to any square not already visited, and only Black is limited to making knight moves.