Since both the problem statement and the winning condition are given in terms of divisibility by $17$, it is sufficient to consider the numbers modulo $17$. In the beginning, all the remainders are different from zero and Alice wins if the sum modulo $17$ becomes zero.

The moves $a \mapsto a^2$ and $b \mapsto b^3$ turn remainders into powers of the original nonzero values. Therefore, Fermat's little theorem can be applied. For $a \not\equiv 0 \pmod{17}$ and the prime number $17$, one has
$$a^{16} \equiv 1 \pmod{17}.$$
So if Alice squares the same number $a$ four times in a row, then the remainder $1$ modulo $17$ is always obtained. Bob cannot do anything about it, because if Bob raises this number to the third power $k$ times, we get a result of
$$a^{2 \cdot 3^k} = 16^{3^k} \equiv 1^{3^k} = 1 \pmod{17}.$$
The timing of Bob's moves does not matter, as the order of the factors in the exponent does not change anything.

Therefore, Alice can make all the remainders equal to $1$ by squaring each number four times. Then of course the sum is $17 \cdot 1 \equiv 0 \pmod{17}$ and Alice has won.