Normal slitherlink rules apply.
The completed grid:
(The little grey marks were just for bookkeeping.)
I'm afraid I didn't keep a detailed record of the solution path.
I started at the corners for obvious reasons, filled in a lot of the top right, extended a bunch of parallel things leftward from the bottom right, and then the rest was fairly straightforward.
Gareth McCaughan got the answer first and got his answer accepted (upvote him too!), but I will be the one to supply the deductions:
Chapter I. The Basic Deductions
Chapter II. Expansion from the Top Right
Chapter III. Invigoration of the Bottom Right
Chapter IV. Continued Extension
Chapter V (Finale). Encounter with the Twos and Completion of the Loop
Now, look at the 2 in Row 9 Column 4 and suppose that its right side were part of the loop. There will be two cases that proceed from here: the “empty” side of the three pointing upward or rightward. The upward case can be divided further into two cases, but however all three cases lead to contradictions involving 2's that cannot be fulfilled:
(In fact, as @aschepler notes, rather than considering those three cases, you can just continue deducing starting from the 2 in Row 8 Column 4 until you reach a contradiction; it becomes similar to the first case depicted above.)
Therefore the “empty” side of the three must be the upper side. This allows us to utilize three simple arguments repeatedly to arrive at the completed loop: first, that there cannot be more than one closed loop; second, that the 2 cannot have more than three sides that are part of the loop; and third, that sometimes an open end is forced to extend in only one direction.