The 1996 NCAA Division III men's ice hockey tournament was the culmination of the 1995–96 season, the 13th such tournament in NCAA history. It concluded with Middlebury defeating RIT in the championship game 3–2. All Quarterfinals matchups were held at home team venues, while all succeeding games were played in River Falls, Wisconsin.

This article relies largely or entirely on a single source. (March 2024)

The following teams qualified for the tournament. There were no automatic bids, however, conference tournament champions were given preferential consideration. No formal seeding was used while quarterfinal matches were arranged so that the road teams would have the shortest possible travel distances.

The tournament featured three rounds of play. In the Quarterfinals, teams played a two-game series where the first team to reach 3 points was declared a winner (2 points for winning a game, 1 point each for tying). If both teams ended up with 2 points after the first two games a 20-minute mini-game used to determine a winner. Mini-game scores are in italics. Beginning with the semifinals all games became Single-game eliminations. The winning teams in the semifinals advanced to the national championship game with the losers playing in a Third Place game. The teams were seeded according to geographic proximity in the quarterfinals so the visiting team would have the shortest feasible distance to travel.

Note: * denotes overtime period(s)
Note: Mini-games in italics