Let the rational number be p/q. Write the condition as $(n+5)(2-p/q)(4-2p/q+p^2/q^2)=75$ and now casework.

Let the the above be equal to some cube of a rational number, $r$. We note that we can express the above as $r=8+\frac{-60}{n+5}$. Subtracting and working some factoring magic we see that if $r=w^3$ for some real $w$ then $(w-2)(w^2+2w+4)=\frac{-60}{n+5}$. Not sure where to proceed from here.... maybe some mod math?

Let $\frac{8n-25}{n+5}=\frac{p^3}{q^3},\ (1)$ where $n,p,q$ are integers with $q\ne 0$ and $(p,q)=1.$ Then $(1)$ is equivalent to $(n+5)(2q-p)(p^2+2pq+4q^2)=65q^3.$ Therefore $n+5=kq^3,$ where $k$ is an integer. Thus $k(2q-p)(p^2+2pq+4q^2)=65=5\cdot 13.$ But $p^2+2pq+4q^2=(p+q)^2+3q^2\ge 3.$ Therefore, we have the following cases: (i) $(p+q)^2+3q^2=5;$ this is impossible in integers. (ii) $(p+q)^2+3q^2=13;$ this implies that $q=\pm 2$ and $p+q=\pm 1.$ We must also have $2q-p=\pm 5$ or $\pm 1.$ The only acceptable solutions in integers are $(p,q)=(-1,2),(1,-2).$ In both cases $n=3.$ (iii) $(p+q)^2+3q^2=65;$ this is impossible in integers.

\[\frac{8n - 25}{n + 5} = 8 - \frac{65}{n + 5}\]Let $\frac{p^3}{q^3} = 8 - \frac{65}{n + 5}$ such that $p \in \mathbb{Z}, q \in \mathbb{Z}^+$ and $\gcd(p, q) = 1$. \[\iff 65q^3 = (n+5)(8q^3 - p^3) = (2q - p)(p^2 + 2pq + 4q^2)(n + 5)\]Since, $\gcd(p, q) = 1$, \[8q^3 - p^3 \equiv -p^3 \not\equiv 0\pmod{q}\]Therefore, $n + 5 = kq^3$ and $k(2q - p)(4q^2 + 2pq + p^2) = 65$, $k\in\{\pm 1, \pm 5, \pm 13, \pm 65\}$, since, $8q^3 - p^3 \in\mathbb{Z}$. Note that $4q^2 + 2pq + p^2 = (p + q)^2 + 3q^2$. Also, note that $n^2\mod{3} \in \{0, 1\}$. Therefore, $(p + q)^2 + 3q^2 \geq 3q^2 \geq 3$. $(p + q)^2 + 3q^2 \equiv (p + q)^2 \equiv 0, 1\pmod{3}$, thus $4q^2 + 2pq + p^2 =13$. Thus, $k(2q - p) = 5, k\in\{\pm 1,\pm 5\}$. Note that if $(k, p, q) = (l, a, b)$ is a solution then so is $(k, p, q) = (-l, -a, -b)$. However, we are only concerned with $\tfrac{p}{q}$, so it suffices to check the cases where $k \ge 0$. Taking cases: $k = 1$, $2q - p = 5$, thus, $p^2 + 5p + 4 = 0 \implies p \in \{-1, -4\}$ $\implies (p, q)\in \{(-1,2), (-4,\tfrac12)\}$. Therefore, $(p, q) = (-1, 2)$, $k = 5$, $2q - p = 1$, thus,$p^2 + p - 4 = 0 \implies p\not\in\mathbb{Z}$. \[\text{Thus, } -\frac{1}{8} = \frac{p^3}{q^3} = 8 - \frac{65}{n + 5} \implies \frac{65}{n + 5} = \frac{65}{8} \implies n = 3\]. Thus, $\boxed{n = 3}$ is the only integer solution.