if $n=9$ we have to check $(4,3,2),(5,3,1),(6,2,1)$ if $n=8$ we have to check $(4,3,1),(5,2,1)$ if $n=7$ we have to check $(4,2,1)$ if $n=6$ we have to check $(3,2,1)$ None of them work For $n=10$ take $(7,2,1)$

Let $gcd(a, b, c, n) = d$ then the first equation turns into $n' = x + y + z$ where $x, y, z$ are coprime. Having $(a + b)(b + c)(c + a) = m^3$ we can turn it into $(x + y)(y + z)(z + x) = m'^3 \implies (n' - z)(n' - x)(n' - y) = m'^3$. Then we have $$n' - z = k$$$$n' - x = k^2$$$$n' - y = h^3$$. As we are looking for the least possible value of $n$ then $k, h$ should be prime numbers. Adding all these up we get $$2n' = k + k^2 + h^3$$As $k$ and $h$ are prime $h$ should be even to have the sum even. As we are looking for the least possible value then $k$ should be $3$. Then $n' = 10$ and $x, y, z$ are $7, 2 ,1$. To have the smallest $n$ $d = 1$ and $n = 10$. We can't check cases where one of these factors is equal to $1$ because the numbers are distinct. And we can't check cases where any $2$ factors have the same divisor (like $k^2, h^2, kh$) because the sum of these should be even and for this, we should have $k = h = 2$ which contradicts to our conditions on $a, b$ and $c$. We have $1$ case left where each of these factors is a perfect cube. $2n' = k^3 + h^3 + p^3$ but as $k = 2$ then $h = 3$ and $p = 5$ gives as a greater n then the answer is $\boxed{10}$