Is there any generalization of the hyperarithmetical hierarchy using the analytical hierarchy to formulas belonging to third-order logic and above? As I understand, hyperarithmetical sets are defined according to the analytical hierarchy, that is, second-order-logic formulas. There is a generalization of hyperarithmetic theory named α-recursion theory. Do this extension generalizes the hyperarithmetical sets to sets definable with any arbitrary higher-order logic? 
NOTE: I know that α-recursion theory extends the hyperarithmetical sets from ${\omega}_1^{CK} $ to any admissible ordinals, but I have no clue what that means or what is the relationship between ordinals and the number N in Nth-order logic).
My question in the end is motivated by the fact that the hyperarithmetical sets can be defined using infinitely iterated Turing jumps, and I wonder if this can be generalized to sets defined by any arbitrary formula (by arbitrary I mean using  any arbitrary higher-order logic formula).
 A: Asaf pointed you to some resources about the main question you asked, about the hierarchy of $\Sigma^2_n$ sets. I'm afraid I can't offer any further assistance on that, but since you also mentioned $\alpha$-recursion and hyperarithmetic but weren't clear on what they were, I thought I could at least give you a brief outline.
Let's begin by reviewing some basic definitions. We obtain the arithmetical hierarchy by finite iterations of the Turing jump operator, starting with the empty set. The $\Sigma^0_n$ sets are those which are recursively enumerable in $0^{(n-1)}$, the $\Pi^0_n$ sets those which are co-recursively enumerable in $0^{(n-1)}$, and the $\Delta^0_n$ sets those which are recursive in $0^{(n-1)}$ (for $n \geq 1$). A set is arithmetical just in case it is $\Sigma^0_n$ for some $n \in \mathbb{N}$.
Now, suppose we want to go further and define the $\omega$-jump, $0^{(\omega)}$. This is recursively isomorphic to the first order theory of the natural numbers, $Th(\mathbb{N})$. By Tarski's theorem $Th(\mathbb{N})$ can't be defined by any arithmetical sentence, so $0^{(\omega)}$ can't be arithmetical. If we want to talk about the definability of sets like $0^{(\omega)}$ (which is, remember, just a subset of the natural numbers) then we'll need to go beyond the arithmetical hierarchy and into hyperarithmetic.
The hyperarithmetical sets are obtained just as the arithmetical ones are, by iterating the Turing jump. It's just that now we iterate not along natural numbers but along recursive ordinals. An ordinal $\alpha$ is a recursive ordinal iff there exists a recursive relation $<_c$ on $X \subseteq \mathbb{N}$ which is a wellordering of order-type $\alpha$. And a set $Y \subseteq \mathbb{N}$ is hyperarithmetical iff $Y$ is recursive in $0^{(\delta)}$ for some recursive ordinal $\delta$.
You mention the analytical hierarchy at the beginning of your question in connection with hyperarithmetic, but in fact the hyperarithmetic sets are very low down in the analytical hierarchy: they are precisely the $\Delta^1_1$ sets. In terms of definability we can go much further with the analytical hierarchy than we can with hyperarithmetic. For instance, the set $\mathcal{O}$ of notations for recursive ordinals is a $\Pi^1_1$ complete set. The supremum of the ordinals which have notations in $\mathcal{O}$ is $\omega_1^{CK}$, the first non-recursive ordinal. So $\omega_1^{CK}$ plays the same role for the hyperarithmetical sets as $\omega$ does for the arithmetical sets.
This brings us to the next generalisation, $\alpha$-recursion. What happens when instead of $\omega_1^{CK}$ we consider some arbitrary ordinal $\alpha$? Clearly not just any $\alpha$ will do: it must have some of the nice closure properties enjoyed by $\omega$ and $\omega_1^{CK}$. This leads to the notion of an admissible ordinal: an ordinal $\alpha$ such that $L_\alpha$ is a transitive model of Kripke–Platek set theory. $\omega$ and $\omega_1^{CK}$ are the first two admissible ordinals. Given an admissible ordinal $\alpha$, $\alpha$-recursive is defined as $\Delta^0_1$ over $L_\alpha$ while $\alpha$-recursively enumerable is $\Sigma^0_1$ over $L_\alpha$. The definitions of relative recursiveness and the jump operation are more complex so there I'll leave you to Sacks's book.
A: I don't know enough to give a full answer, but I did ask [at least] two prominent set theorists about this a year ago.
I was referred to papers of Abraham & Shelah about third-order well-orderings of the real numbers:


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*A $\Delta^2_2$ Well-Order of the Reals And Incompactness of $L(Q^{MM})$. Annals of Pure and Applied Logic 59 (1993) 1--32.

*Martin's axiom and $\Delta^2_1$ well-ordering of the reals. Archive for Mathematical Logic, 35 (1996) 287--298.

*Coding with ladders a well-ordering of the reals. The Journal of Symbolic Logic, 67, Number 2 (2002) 579--597.
