The final answer is
Fill in some lines according to basic Slitherlink deduction techniques. Note that the squares diagonal to the '0' cannot be part of the region that the '0' themselves are part of (because of the no 2 x 2 rule). Hence, we can draw lines between those squares as well:
Then, focus on the middle left where the three '1's are. The loop cannot be on the left for both '1's in R5C1 and R6C1 as it would cause 2 x 2 area in R5:6C1:2. If the loop runs through the top of R5C1, then part of the loop segment must pass through the bottom of R6C1, which will again create a 2 x 2 area in R5:6C1:2. Thus, part of the loop segment must be on the right of R5C1.
As pointed out by @fljx in the comments, there is a particular arrangement at this point where the loop may still possibly pass through the top of R6C2. However, this will soon reach a contradiction as the '2' clue would not be satisfied (or there would be a 2 x 2 region in R4:5C2:3).
So, this means that the loop segment to the right of R5C1 must go down instead.
Then, note that the top left must be 'L' tetromino. Therefore, the square at R3C2 must not be a part of a 'L' tetromino. But, there is no way to fit an 'S' or a 'T' tetromino involving R3C2, so R3C2 must be part of an 'I' tetromino. This 'I' tetromino cannot lie horizontally (otherwise, the square at R2C4 is isolated), so it must lie vertically. Thus, the square R4C1 will not be a part of this region and is also a part of an 'I' tetromino.
From there, we can make some deductions towards the bottom and middle left of the puzzle.
At this point, if the loop above R3C2 were to go down, then we get the following contradiction where there is a 2 x 2 area at R2C6-R3C7 (or two isolated squares)
So that loop doesn't go down but must go right. This creates a few more knock-on deductions and we can then get to here.
Then, note that the square at R1C8 cannot be part of an 'L' tetromino (as it would share an edge with another 'L' tetromino in the same region. It can't be part of an 'S' or 'T' tetromino, so it must be part of an 'I' tetromino. This 'I' tetromino must lie horizontally (as it would create an isolated square at R3C7 if it were to lie vertically instead).
Then, the loop above R3C9 cannot go to the right as it would create an isolated square at either R2C9 or R1C12 instead. So, it must go down, and from the deductions that follow, we can fill in some tetromino shapes as well.
If the loop segment above R4C8 were to go down, then the loop segment to the right of R3C9 must also go down. This will create isolated squares at R4C9 and R5C9, so the loop segment above R4C8 must go right. This leads us to here.
Then, the square at R6C8 cannot be part of a 'T' tetromino. It cannot be part of an 'I' as well, as it would leave isolated squares at R5C9 and R5C10. So, it must be part of an 'L'. From there, we can connect the connections in the middle right area and fill in the corresponding tetromino shapes.
Now, we can resolve the '3' at the bottom right as well, as there is only one way to connect the links there to create an outside region of area 4.
Then, look at the squares R7C4, R8C4, R8C5. These three are all linked together and will either form an 'S' tetromino or an 'L' tetromino. It can't be an 'S' as it would create an isolated square at R9C4. So, it must be an 'L' and we can then fill in a few more tetrominoes on the bottom right.
Finally, we note that the loop segment to the right of R8C6 cannot go down as it would create an isolated loop. So, it must go left and connect up the loop segment that is above R9C4. We can then easily sort out the final connections to get the final answer.