Besicovitch has proved the following related interesting result:

Consider an integer $n\gt 1$ and distinct prime numbers $p_1,p_2,\ldots ,p_k.$ Then the field $F=\mathbb Q (\sqrt[n]{p_1},\ldots ,\sqrt[n]{p_k})$ has dimension $n^k$ over $\mathbb Q$ .
More precisely, a $\mathbb Q$-basis of that field $F$ is given by the radicals $$\sqrt[n]{p_1^{m_1}\ldots p_i^{m_i} \ldots p_k^{m_k} } \quad (\; 0\leq m_i \lt n \quad , \quad 1\leq i\leq k )$$

(The case $n=2$ is a classical chestnut in Galois theory.)
This does not answer the OP's question but at least assures us that, for example, $$\sqrt[3]{900}+\sqrt[3]{36}+ \sqrt[3]{15}+\sqrt[3]{150} \notin \mathbb Q$$
which is not so simple to check directly.

I have the pessimistic feeling that there is no very satisfactory general answer to the question "when does the sum $\sqrt[n_1]{a_1}+ \sqrt[n_2]{a_2}+...+\sqrt[n_k]{a_k}$ have degree $n_1 n_2 ...n_k$", but I'd love to be shown wrong.

Bibliography: Besicovich's original article is: Abram S. Besicovitch, "On the linear independence of fractional powers of integers", Journal of the London Mathematical Society 15 (1940), 3-6.

Here is a more recent and accessible proof : Ian Richards, "An application of Galois theory to elementary arithmetic", Advances in Mathematics 13 (1974), 268-273. 13 (1974), 268-273.

2 added numerical example and pessimistic comment

Besicovitch has proved the following related interesting result:

Consider an integer $n\gt 1$ and distinct prime numbers $p_1,p_2,\ldots ,p_k.$ Then the field $F=\mathbb Q (\sqrt[n]{p_1},\ldots ,\sqrt[n]{p_k})$ has dimension $n^k$ over $\mathbb Q$ .
More precisely, a $\mathbb Q$-basis of that field $F$ is given by the radicals $$\sqrt[n]{p_1^{m_1}\ldots p_i^{m_i} \ldots p_k^{m_k} } \quad (\; 0\leq m_i \lt n \quad , \quad 1\leq i\leq k )$$

(The case $n=2$ is a classical chestnut in Galois theory.)
This does not answer the OP's question but at least assures us that, for example, $$\sqrt[3]{900}+\sqrt[3]{36}+ \sqrt[3]{15}+\sqrt[3]{150} \notin \mathbb Q$$
which is not so simple to check directly.

I have the pessimistic feeling that there is no very satisfactory answer to the question "when does the sum $\sqrt[n_1]{a_1}+ \sqrt[n_2]{a_2}+...+\sqrt[n_k]{a_k}$ have degree $n_1 n_2 ...n_k$", but I'd love to be shown wrong.

Bibliography: Besicovich's original article is: Abram S. Besicovitch, "On the linear independence of fractional powers of integers", Journal of the London Mathematical Society 15 (1940), 3-6.

Here is a more recent and accessible proof : Ian Richards, "An application of Galois theory to elementary arithmetic", Advances in Mathematics 13 (1974), 268-273. 13 (1974), 268-273.

1

Besicovitch has proved the following related interesting result:

Consider an integer $n\gt 1$ and distinct prime numbers $p_1,p_2,\ldots ,p_k.$ Then the field $F=\mathbb Q (\sqrt[n]{p_1},\ldots ,\sqrt[n]{p_k})$ has dimension $n^k$ over $\mathbb Q$ .
More precisely, a $\mathbb Q$-basis of that field $F$ is given by the radicals $$\sqrt[n]{p_1^{m_1}\ldots p_i^{m_i} \ldots p_k^{m_k} } \quad (\; 0\leq m_i \lt n \quad , \quad 1\leq i\leq k )$$

(The case $n=2$ is a classical chestnut in Galois theory.)

Bibliography: Besicovich's original article is: Abram S. Besicovitch, "On the linear independence of fractional powers of integers", Journal of the London Mathematical Society 15 (1940), 3-6.

Here is a more recent and accessible proof : Ian Richards, "An application of Galois theory to elementary arithmetic", Advances in Mathematics 13 (1974), 268-273. 13 (1974), 268-273.