Let $X$ be an Banach space, and $X^*$ the space of linear functionals on $X$. The dual of $X^*$ is called the bidual, and if the bidual $X^{**}=X$, we say that $X$ is a reflexive space. It is well known that the $L^p$-spaces ($1<p<\infty$) are reflexive.
Now, let us define the tridual to be the dual of the bidual, $X^{***}$. Are there spaces $X$ such that $X^{***}=X$? What about "reflexivity" with respect to n-duals? Does this have any application?
I am not sure what you mean by the equality $X^{***} = X$, if you mean that $X$ and $X^{***}$ are isometrically isomorphic then the answer is no, it does not imply reflexivity.
There is a classical example $J$ of a Banach space constructed by R. C. James (the James space) with the property that $J$ is isometrically isomorphic to $J^{**}$, yet $J$ is not reflexive:
Take $X = J\oplus_2 J^*$. Then $X$ is not reflexive as it contains a non-reflexive subspace $J$ but
$X^{***} \equiv J^{***} \oplus_2 J^{****}\equiv J^* \oplus_2 J\equiv J\oplus_2 J^* = X,$
where by $\equiv$ I mean the existence of an isometric isomorphism. Of course, this answers your question in negative for all $n$ as well.