Assume that the seats are all numbered sequentially from 1 to 8 and placed around the circular table in order in clockwise direction.


Next assume that the freshmen occupies the chair #1.


Then you may think, as a first approximation, that each of the three groups (3 seniors / 2 juniors / 2 sophomores) represents one object.


Then you have 3 objects, and there are 3! = 1*2*3 = 6 permutations to order them.       (1)


Inside of each group you have 3! = 6 ways to order seniors,                             (2)

                              2! = 2 ways to order juniors  and                         (3)

                              2! = 2 ways to order sophomores,                          (4)

and these interior orderings are independent.


Hence, the final answer is  

(6 permutations of (1)) * (6 permutations of (2) ) * (2 permutations of (3) ) *(2 permutations of (4) ) = 6*6*2*2 = 144.


Answer.  There are 144 ways (144 circular permutations) to do it.