Without loss of generality, assume $a \leq b \leq c$.

If $c \geq 6$, then $c!$ is divisible by $2^4 = 16$, and $a!$ and $b!$ are also divisible by $2$ for $a, b \geq 2$. Thus, $a! + b! + c!$ is divisible by $2$, but for $c \geq 6$, $c!$ is divisible by $8$ and higher powers of $2$.

Let us check small values of $c$:

If $c = 1$:
$a! + b! + 1! = 2^n$
But $a, b \leq 1$, so $a = b = 1$.
$1! + 1! + 1! = 1 + 1 + 1 = 3$, which is not a power of $2$.

If $c = 2$:
$a! + b! + 2! = a! + b! + 2 = 2^n$
$a, b \leq 2$
Try $a = 1, b = 1$: $1 + 1 + 2 = 4 = 2^2$
So $n = 2$ is a solution with $(a, b, c) = (1, 1, 2)$ (and permutations).

Try $a = 1, b = 2$: $1 + 2 + 2 = 5$
$a = 2, b = 2$: $2 + 2 + 2 = 6$
No other powers of $2$.

If $c = 3$:
$a! + b! + 6 = 2^n$, $a, b \leq 3$
$a = 1, b = 1$: $1 + 1 + 6 = 8 = 2^3$
So $n = 3$ is a solution with $(a, b, c) = (1, 1, 3)$ (and permutations).

$a = 1, b = 2$: $1 + 2 + 6 = 9$
$a = 1, b = 3$: $1 + 6 + 6 = 13$
$a = 2, b = 2$: $2 + 2 + 6 = 10$
$a = 2, b = 3$: $2 + 6 + 6 = 14$
$a = 3, b = 3$: $6 + 6 + 6 = 18$
No other powers of $2$.

If $c = 4$:
$4! = 24$
$a, b \leq 4$
$a = 1, b = 1$: $1 + 1 + 24 = 26$
$a = 1, b = 2$: $1 + 2 + 24 = 27$
$a = 1, b = 3$: $1 + 6 + 24 = 31$
$a = 1, b = 4$: $1 + 24 + 24 = 49$
$a = 2, b = 2$: $2 + 2 + 24 = 28$
$a = 2, b = 3$: $2 + 6 + 24 = 32 = 2^5$
So $n = 5$ is a solution with $(a, b, c) = (2, 3, 4)$ (and permutations).

$a = 2, b = 4$: $2 + 24 + 24 = 50$
$a = 3, b = 3$: $6 + 6 + 24 = 36$
$a = 3, b = 4$: $6 + 24 + 24 = 54$
$a = 4, b = 4$: $24 + 24 + 24 = 72$

If $c = 5$:
$5! = 120$
$a, b \leq 5$
$a = 1, b = 1$: $1 + 1 + 120 = 122$
$a = 1, b = 2$: $1 + 2 + 120 = 123$
$a = 1, b = 3$: $1 + 6 + 120 = 127$
$a = 1, b = 4$: $1 + 24 + 120 = 145$
$a = 1, b = 5$: $1 + 120 + 120 = 241$
$a = 2, b = 2$: $2 + 2 + 120 = 124$
$a = 2, b = 3$: $2 + 6 + 120 = 128 = 2^7$
So $n = 7$ is a solution with $(a, b, c) = (2, 3, 5)$ (and permutations).

$a = 2, b = 4$: $2 + 24 + 120 = 146$
$a = 2, b = 5$: $2 + 120 + 120 = 242$
$a = 3, b = 3$: $6 + 6 + 120 = 132$
$a = 3, b = 4$: $6 + 24 + 120 = 150$
$a = 3, b = 5$: $6 + 120 + 120 = 246$
$a = 4, b = 4$: $24 + 24 + 120 = 168$
$a = 4, b = 5$: $24 + 120 + 120 = 264$
$a = 5, b = 5$: $120 + 120 + 120 = 360$

For $c \geq 6$, $c!$ is divisible by $16$, and $a! + b!$ is much smaller than $c!$, so $a! + b! + c!$ cannot be a power of $2$.

Therefore, the possible values of $n$ are $2, 3, 5, 7$.