It is clear that if $n \ge 3$, $n! \nmid 3$, that is $n! \ne 3k$, for some natural number $k$. But then $2^{n!} = 2^{3k} = 8^k \equiv 1 \pmod{7}$. Analogously modulo 9: if $n \ge 3$ we have that $n! \nmid 6$, that is $n! \ne 6l$, for some natural $l$. But then $2^{n!} = 2^{6l} = 64^l \equiv 1 \pmod{9}$.

It's easy to see, that the cubes of integer numbers are equal $0$ or $\pm 1$ modulo $7$ and modulo $9$. Without loss of generality, we may assume that $a \ge b \ge c$.

Case 1. $a \ge 3$ and $b \ge 3$. If $c \ge 3$, then $2^{a!} + 2^{b!} + 2^{c!} \equiv 3 \pmod{7}$ – is not a cube of an integer.
If $c=2$ then $2^{a!} + 2^{b!} + 2^{c!} \equiv 1+1+4 \equiv -1 \pmod{7}$ – can be a cube of an integer.
Check modulo 9. $2^{a!} + 2^{b!} + 2^{c!} \equiv 1+1+4 \equiv 6 \pmod{9}$ – is not a cube of an integer.
If $c=1$ then $2^{a!} + 2^{b!} + 2^{c!} \equiv 1+1+2 \equiv 4 \pmod{7}$ – is not a cube of an integer.

Case 2. $a \ge 3 \Rightarrow b \le 2$. Hence we have three cases.
If $b=c=2$ $2^{a!} + 2^{b!} + 2^{c!} \equiv 1+4+4 \equiv 2 \pmod{7}$ – is not a cube of an integer.
If $b=2, c=1$ $2^{a!} + 2^{b!} + 2^{c!} \equiv 1+4+2 \equiv 0 \pmod{7}$ – can be a cube of an integer.
Then analogously modulo 9 $2^{a!} + 2^{b!} + 2^{c!} \equiv 1+4+2 \equiv 7 \pmod{9}$ – is not a cube of an integer.
If $b=c=1$ $2^{a!} + 2^{b!} + 2^{c!} \equiv 1+2+2 \equiv 5 \pmod{7}$ – is not a cube of an integer.

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Case 3. $2 \ge a \ge b \ge c$, here we have 4 cases overall, from which we simply find a single answer in a simple overview.