Step 2 causes at most $n$ repetitions since each repetition other than the last one marks a node which wasn't previously marked. Since there are only $n$ nodes and one of them is marked in Step 1, there are at most $n-1$ repetitions which mark a new node, and one more which doesn't (and so moves to Step 4).
In Step 3, we go over all nodes in $G$ ($n$ nodes), and check all its neighbors ($n-1$ neighbors), so there are at most $n(n-1)$ repetitions of the inner part.
As for how many steps does this take when implemented on a Turing machine, this depends on the implementation. Outside of introductory courses on theory of computation, we almost never consider Turing machines, preferring instead to work with more intuitive models such as the RAM machine. Therefore mastering the intricacies of Turing machine behavior is not so important.