1. Let $p = \sqrt[5]{m + \sqrt{n}}$ and $q = \sqrt[5]{m - \sqrt{n}}$. We are given that $x = p + q$ is a real root of the polynomial $x^5 - 10x^3 + 20x - 40$.

2. Consider the polynomial $x^5 - 10x^3 + 20x - 40$. We need to express $x^5$ in terms of $p$ and $q$.

3. First, note that:
$$p^5 = m + \sqrt{n} \quad \text{and} \quad q^5 = m - \sqrt{n}$$
Therefore,
$$p^5 + q^5 = (m + \sqrt{n}) + (m - \sqrt{n}) = 2m$$

4. Next, we need to find $pq$:
$$pq = \sqrt[5]{(m + \sqrt{n})(m - \sqrt{n})} = \sqrt[5]{m^2 - n}$$
Given $m^2 - n = 32$, we have:
$$pq = \sqrt[5]{32} = 2$$

5. Now, we express $x^5$ in terms of $p$ and $q$:
$$x = p + q$$
$$x^5 = (p + q)^5 = p^5 + q^5 + 5pq(p^3 + q^3) + 10p^2q^2(p + q)$$
Since $p^5 + q^5 = 2m$ and $pq = 2$, we need to simplify the expression:
$$x^5 = 2m + 5pq(p^3 + q^3) + 10p^2q^2(p + q)$$

6. We know:
$$p^3 + q^3 = (p + q)(p^2 - pq + q^2)$$
$$p^2 + q^2 = (p + q)^2 - 2pq = x^2 - 4$$
$$p^3 + q^3 = x(x^2 - 3pq) = x(x^2 - 6)$$

7. Substituting back, we get:
$$x^5 = 2m + 5 \cdot 2 \cdot x(x^2 - 6) + 10 \cdot 4 \cdot x$$
$$x^5 = 2m + 10x^3 - 60x + 40x$$
$$x^5 = 2m + 10x^3 - 20x$$

8. Given that $x$ is a root of $x^5 - 10x^3 + 20x - 40 = 0$, we substitute:
$$x^5 - 10x^3 + 20x - 40 = 0$$
$$2m + 10x^3 - 20x - 10x^3 + 20x - 40 = 0$$
$$2m - 40 = 0$$
$$2m = 40$$
$$m = 20$$

9. Using $m = 20$ in $m^2 - n = 32$:
$$20^2 - n = 32$$
$$400 - n = 32$$
$$n = 368$$

10. Therefore, $m + n = 20 + 368 = 388$.

The final answer is $\boxed{388}$