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

If $c\ge 6$, then $c!$ is divisible by $2^4=16$, and $a!$ and $b!$ are also divisible by $2$ for $a,b\ge 2$. Thus, $a!+b!+c!$ is divisible by $2$, but for $c\ge 6$, $c!$ 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 $a,b\le 1$, so $a=b=1$. $1!+1!+1!=1+1+1=3$, which is not a power of $2$.

If $c=2$: $a!+b!+2!=a!+b!+2=2^n$ $a,b\le 2$ 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 $c=3$: $a!+b!+6=2^n$, $a,b\le 3$ $a=1,b=1$: $1+1+6=8=2^3$ 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 $c=4$: $4!=24$ $a,b\le 4$ $a=1,b=1$: $1+1+24=26$ $a=1,b=2$: $1+2+24=27$ $a=1,b=3$: $1+6+24=31$ $a=1,b=4$: $1+24+24=49$ $a=2,b=2$: $2+2+24=28$ $a=2,b=3$: $2+6+24=32=2^5$ 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 $c=5$: $5!=120$ $a,b\le 5$ $a=1,b=1$: $1+1+120=122$ $a=1,b=2$: $1+2+120=123$ $a=1,b=3$: $1+6+120=127$ $a=1,b=4$: $1+24+120=145$ $a=1,b=5$: $1+120+120=241$ $a=2,b=2$: $2+2+120=124$ $a=2,b=3$: $2+6+120=128=2^7$ 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 $c\ge 6$, $c!$ 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 $2,3,5,7$.