In a legal position, how many times can the king and its surrounding squares be attacked? There is a total of 9 possible squares to be attacked-the square the king is on and the eight squares surrounding it. The final tally is the added total of the number of times each of the nine squares is attacked.
Adding on to @Brian Tower's answer as anticipated, here is an improved position with 55 attacks, using double check and White's full arsenal. Black last moved a piece to e7 that White's rook captured with checkmate.
[FEN "B7/4R3/3Q1Q2/2Q3Q1/2N1k1K1/1R4Q1/1BNQNQN1/8 w - - 0 1"]
However, it is also possible for both sides to attack under the OP's loose ruleset. I have found a lower bound of 83 attacks by White and Black, with eight of them coming from the Black king itself.
[FEN "8/3Nnn2/1bnQqQb1/1r4Q1/1nq1k1qN/1R4Q1/1BNQqQB1/2K1Nn2 w - - 0 1"]
If you are wondering of all of those pawns really could have promoted, I present evidence that it is possible.
[FEN ""] 1. a4 b5 2. h4 g5 3. c4 d5 4. f4 e5 5. cxd5 bxa4 6. fxe5 g4 7. Rh3 gxh3 8. h5 Nf6 9. h6 Rg8 10. e6 Rg7 11. hxg7
Finally, just for fun, here is 88 attacks by one side in an illegal position. Any piece can be color swapped to produce a result for both sides.
[FEN "8/2NNNNN1/1NNQQQNN/1NQ3QN/1NQ1K1QN/1NQ3QN/1NNQQQNN/2NNNNN1 w - - 0 1"]
There seems to be some confusion understanding the problem, so I will post an answer, probably not the maximum, and then somebody better than me can post the best solution.
I've promoted 7 pawns to queens and one pawn to a knight.
[fen "R7/7b/3qnq2/2q3q1/2n1K1n1/2q3q1/3q1q3/7k w - - 0 1"]
Each of the 8 queens attacks 5 squares around the king = 40 Each of the 3 knight attacks 2 squares around the king = 6 The bishop attacks the king and 1 square around the king = 2
Total = 48