DFA accepting strings with at least three occurrences of three consecutive 1's
No, you don't need NFA (although it is equivalent to DFA) or PDA or Turing machine to solve this problem. It can be easily done with a DFA.

Lets number the states in given DFA in clockwise fashion starting from start state as Q0 to final state as Q9. First of all, it being a DFA, Q9 should have a self loop with label 1 as all possible transitions should be shown from all the states in a DFA. 

Now we need to add two states, lets name them Qa and Qb. The changes in the transitions to be made are as follows:-


Instead of having a 1 transition from Q2 to Q3, make a 1 transition from Q2 to Qa.
Instead of having a 1 transition from Q5 to Q6, make a 1 transition from Q5 to Qb.
Make a new 1 transition from Qa to Qb and a 0 transition from Qa to Q3.
Make a new 1 transition from Qb to Q9 and a 0 transition from Qb to Q6.


These changes in the provided DFA will allow it to satisfy given conditions. It simply counts input substring 1111 as two consecutive 111 and skips a step (of three intermediate states), increasing the count be 2.