The number $\sqrt{a}+\sqrt{b}$ cannot be an integer if $a,b$ are integers such that $\sqrt{b}$ is not an integer. (In fact, this is true for any number of square roots, and I believe even for cube roots, etc.) Therefore $\sqrt[n]{\sqrt{a}+\sqrt{b}}$ cannot be an integer either, for any positive integer $n$.
What about if we sum it with the conjugate? That is, do there exist positive integers $n\geq 2,a,b$ such that $\sqrt{b}$ is not an integer but $\sqrt[n]{\sqrt{a}+\sqrt{b}}+\sqrt[n]{\sqrt{a}-\sqrt{b}}$ is an integer?
Update: As Jyrki Lahtonen points out in the comment, there do exist such integers. What about if we also require that $\sqrt{a}$ is not an integer either?
The answer is yes. Pell's equations solve the problem.
Let
$$A.N'^{2}+1=M'^{2}$$
the generic Pell equation, the solution of which is:
$N’=\frac{(M+N\sqrt{A})^{n}}{2\sqrt{A}}-\frac{(M-N\sqrt{A})^{n}}{2\sqrt{A}}$,
$M’==\frac{(M+N\sqrt{A})^{n}}{2}+\frac{(M-N\sqrt{A})^{n}}{2}$
where $ M$ and $N$ are the basic solutions of $A$
(For example if we choose $A=8$, $N=1$ and $M=3$).
$\sqrt[n]{\sqrt{a}+\sqrt{b}}+\sqrt[n]{\sqrt{a}-\sqrt{b}}=M’$,
$\sqrt[n]{\sqrt{a}+\sqrt{b}}-\sqrt[n]{\sqrt{a}-\sqrt{b}}=N’\sqrt{A}$;
$a=2^{-2(n+1)}\Big((M+N\sqrt{A})^{n}+(M-N\sqrt{A})^{n}\Big)^{2}$,
$b=2^{-2(n+1)}\Big((M+N\sqrt{A})^{n}-(M-N\sqrt{A})^{n}\Big)^{2}$.
Let's take an example: $A=13$, $N=180$ and $M=649$,
we choose $n=4$, then
$\sqrt[4]{\sqrt{a}+\sqrt{b}}+\sqrt[4]{\sqrt{a}-\sqrt{b}}=1419278889601$,
$a=\frac{1053711982714216551873491874429255128997076465231993345538436469128908160121757992166220331730227201}{256}$,
$b=4116062432477408405755827634489277847644829942312474006009517457534797500475617156899298170821200$.
If we choose $A=8$,$M=3$ and $N=1$ and $n=5$, we get:
$\sqrt[4]{\sqrt{a}+\sqrt{b}}+\sqrt[4]{\sqrt{a}-\sqrt{b}}=3363$,
$a=\frac{47370562574818466708936539960450008969}{1024}$,
$b=\frac{5921320321852308338617067495056251121}{128}$,
and so on…