Directly checking yields:
$$p_1! = 2, \quad p_1! + p_2! = 8 = 2^3, \quad p_1! + p_2! + p_3! = 128 = 2^7$$
For $n \ge 4$, $p_1! + p_2! + \dots + p_n! = 128 + 7! + 11! + \dots + p_n!$ is a number in which all summands are divisible by $2^5$, except $p_4! = 7! = 2^4 \cdot 315$, which is divisible by $2^4$ but not by $2^5$. Therefore, this number cannot be a power of 2. If it is a power of another number, then since $2^4$ divides it, it can only be $2^4$ or $x^2$. Let us show that it cannot be a square. Indeed, $2! + 3! + \dots + p_n!$ gives a remainder 2 modulo 3, which is impossible for squares.