The answer is 3. First, notice that the last digit of Magalí's initial number, name it $n_1$, determines the last digit of the following numbers. This is explained in the following table, where we name $n_2 = n_1^2 + 6$, $n_3 = n_2^2 + 6$, $n_4 = n_3^2 + 6$ and $n_5 = n_4^2 + 6$:

For example, if $n_1$ had 4 as its last digit, then $n_2 = n_1^2 + 6$ would have the same last digit as $4 \times 4 + 6 = 22$, which is 2. We omitted the numbers with their last digit being even but different from 2 because they cannot be prime. Now, observe on the table that no matter which prime number Magalí chooses initially, in at most three steps she reaches a number with last digit 0 or 5, and so a multiple of 5. Since this number is greater than 5, it is not a prime number. Therefore, she can never press the ⋆ button more than three times. It is possible that Magalí presses ⋆ three times. Indeed, if she starts with $n_1 = 5$, the following numbers are $n_2 = 5^2 + 6 = 31$, $n_3 = 31^2 + 6 = 967$ and $n_4 = 967^2 + 6$. Since 5, 31 and 967 are prime numbers and $967^2 + 6$ is a multiple of 5, this proves the desired result.