Below I've addressed your specific questions. However, based on your multiple questions about this I think it might be more useful to give a list of good sources, so I'll do that first.
On "gaps" in the constructible universe: Marek/Srebrny, Gaps in the constructible universe. The introduction is very readable and will give you a good sense of what's going on.
On the mastercode hierarchy (and what happens when new reals do appear): Hodes' paper Jumping through the transfinite. This is also closely connected with the study of gaps. Like the paper above, the introduction is a very good read.
On the general structure of $L$: Devlin's book Constructibility. It has a serious error unfortunately, but that error doesn't affect the important results; see this review by Stanley for a summary of the issue (and if you're interested in how to correct it, this paper by Mathias). Ultimately the error is very limited and easily avoided once you know it exists - basically, doubt anything involving a claim about the (aptly named) set theory "BS," but pretty much everything else is correct.
Now, it would seem that every one of these sets could in principle be defined in first order logic without parameters (though I am unsure how this would work in practice)
There's no subtlety here: we first define addition and multiplication of finite ordinals, and now we can use port the usual definitions in $(\mathbb{N}; +,\times)$ of those sets into the set theory context. Indeed, there's a natural way (the Ackermann interpretation) to pass between $L_\omega$ and $(\mathbb{N};+,\times)$, so definability in $L_\omega$ can be reasoned about by proving things in the more familiar setting of definability in arithmetic; e.g. this lets us argue that the Busy Beaver function is indeed in $L_{\omega+1}$.
would a non-constructible real (assuming its existence) be in some sense infinitely complex in that it could not be described in any form whatsoever, either directly, or via some cumulative process?
Certainly not: e.g. $0^\sharp$ is definitely definable (it's $\Delta^1_3$, and in particular is definable in second-order arithmetic) but is not in $L$ (assuming it exists at all). ZFC can't prove that something matching the definition of $0^\sharp$ exists, but it can prove that if it exists then it's not constructible.
Given a particular countable ordinal $\alpha$, can we always find (by which I mean, explicitly describe) a real X with L-rank $\alpha$?
No; for many (indeed, club-many) ordinals $<\omega_1^L$, we have no new reals at that level. Indeed, the $L$-hierarchy is "filled with gaps" - even very long gaps. If you google "gaps in $L$-hierarchy" you'll find a lot of information around this; roughly speaking, an ordinal $\alpha<\omega_1^L$ starts a "long" gap if it is "very" similar to $\omega_1^L$.
In terms of complexity, the reals clearly become more complex as their $L$-rank increases, but is there a way to formalize this precisely?
Well, the obvious one is that if $A$ has $L$-rank greater than that of $B$, then the set $A$ is not definable in the structure $(\mathbb{N}; +,\times, B)$ (that is, arithmetic augmented by a predicate naming the naturals in $B$). In particular $A\not\le_TB$. On the other hand, $A$ might not compute $B$ either (e.g. if $A$ is "sufficiently Cohen generic" over $L_\beta$ then $A$ won't compute any noncomputable real in $L_\beta$ - in particular, it won't compute any real in $L_\beta$ not in $L_{\omega+1}$).
Throughout, by "structure" I mean "countable structure in a computable language." I'm also assuming you're comfortable both with Turing reducibility $\le_T$ - which lets us avoid unnecessary verbiage about machines and oracles - and with the idea of reals coding copies of structures (see SSequence's answer, specifically the $\omega\cdot2$ example).
We begin with the computability side. For $r$ a real, we let $\omega_1^{CK}(r)$ be the smallest ordinal with no $r$-computable copy; equivalently, the supremum of the ordinals which do have $r$-computable copies. For a structure $\mathcal{A}$ we let $\omega_1^{CK}(\mathcal{A})$ be the smallest ordinal not computed by some$^1$ copy of $\mathcal{A}$; precisely, $$\omega_1^{CK}(\mathcal{A})=\min\{\omega_1^{CK}(r):r\mbox{ codes a copy of $\mathcal{A}$}\}.$$
- "$\omega_1^{CK}(r)$" is not how that appears in the literature - rather, you'll see "$\omega_1^r$" - but I strongly prefer it, since it avoids clashing with set-theoretic notation. Note also that we can conflate a real $r\subseteq\omega$ with the structure $\hat{r}$ consisting of the natural numbers with successor and a unary predicate for $r$, and it's easy to check that $\omega_1^{CK}(r)=\omega_1^{CK}(\hat{r})$, so everything lines up nicely.
Next, we look at the admissibility side. For $\alpha$ an arbitrary ordinal, we let $\omega_\alpha^{ad}$ denote the $\alpha$th admissible ordinal: that is, the $\alpha$th ordinal whose corresponding level of $L$ satisfies KP. Note that this definition has nothing to do with computability theory on the face of it (and in fact, doesn't even require $\alpha$ to be countable!). We'll also write "$\omega_1^{ad}(\beta)$" for the first admissible ordinal $>\beta$; in particular, $\omega_1^{ad}(\omega_\alpha^{ad})=\omega_{\alpha+1}^{ad}$.
- Nobody uses this notation, since by Sacks' result it's completely redundant. However, distinguishing at this stage in the game between admissibility concerns and computability concerns is I think very helpful, so I hope you'll forgive me the introduction of soon-to-be-stupid notation.
Now Sacks' result (plus a bit of thought) shows that $$\omega_1^{CK}(\alpha)=\omega_1^{ad}(\alpha)\mbox{ for every countable ordinal $\alpha$}.$$ This is why you never see the "$ad$" notation: it's made completely irrelevant! In particular, "$\omega_\alpha^{CK}$" is just our "$\omega_\alpha^{ad}$."
Moreover, Sacks' result immediately implies that $\omega_1^{CK}(\mathcal{A})$, being the minimum of a set of admissible ordinals,is itself admissible.
Also, via forcing we can make sense of this for even uncountable $\alpha$. But that's really a side issue.
$^1$Note the careful quantification over copies here (and the implicit focus on "optimally simple" copies) in our definition of $\omega_1^{CK}(\mathcal{A})$. This is fundamental: different copies of the same structure can behave very differently, and we need to address this if our definitions are to be interesting at all.
Specifically, we can have very simple structures coded by very complicated reals: e.g. "swapping" $2n$ and $2n+1$ whenever $n\in 0'$ gives a copy of $\omega$ which computes $0'$, and more generally we can get copies of $\omega$ of arbitrarily high complexity. In fact, this (almost) always happens. So in order to say anything interesting, we need to talk about what all copies of a given structure can do.
- Note: this is what Wojowu's comment "Results due to Sacks imply that with such an oracle we can compute all ordinals below $\omega_2^{CK}$, and for suitable choice of this oracle [typo removed] no larger ordinals will be computable with this oracle." Obviously some copies of $\omega_2^{CK}$, when used as oracles, will let us compute a ton of extra junk; the point is that nothing beyond $\omega_2^{CK}$ is necessarily computable from a copy of $\omega_1^{CK}$.
What we're ultimately getting at here is the idea of reducibilities between structures. Here we're looking just at Muchnik (weak) reducibility: $\mathcal{A}\le_w\mathcal{B}$ if every real coding a copy of $\mathcal{B}$ computes some real coding a copy of $\mathcal{A}$. There are other reducibilities too - most immediately, Medvedev (strong) reducibility - but for these sorts of questions we're really in the Muchnik realm, at least for now.
EDIT: An important point here which I think will substantially clarify things is that Muchnik reducibility extends $\le$ - if $\mathcal{A}\ge_w\alpha$ and $\beta<\alpha$ then $\mathcal{A}\ge_w\beta$. In particular this means that $\omega_1^{CK}(\mathcal{A})$ is both the least ordinal without a copy computable from every copy of $\mathcal{A}$, and the supremum of the ordinals which do have copies computable from every copy of $\mathcal{A}$.
EDIT THE SECOND: And here's a way of constructing such a "sufficiently simple" copy of $\omega_1^{CK}$: a copy of $\omega_1^{CK}$ can be computed straightforwardly from Kleene's $\mathcal{O}$, but$^2$ $\mathcal{O}$ is in $L_{\omega_2^{CK}}$ and so every ordinal with a copy computable from $\mathcal{O}$ is $<\omega_2^{CK}$. All of this requires a bit of familiarity with admissible sets and $L_{\omega_1^{CK}}$ in particular; Sacks' book is as usual a good source on the topic.
Best Answer
First off, a correction: you're conflation $L_\alpha$ with $L_\alpha\cap\mathcal{P}(\omega)$. E.g. $L_{\omega_1^{CK}}$ is not the set of hyperarithmetic reals, but $L_{\omega_1^{CK}}\cap\mathcal{P}(\omega)$ is.
Second, for simplicity let's assume $\mathsf{V=L}$. This means that $L_{\omega_1}$ contains every real, so the analytic hierarchy stops well before $L_{\omega_1}\cap\mathcal{P}(\omega)$.
OK, what about smaller admissible ordinals - say, the $\omega_n^{CK}$s for finite $n$ - and the first couple levels of the analytical hierarchy? Well, already in $L_{\omega_2^{CK}}$ we have Kleene's $\mathcal{O}$, and so $L_{\omega_2^{CK}}\cap\mathcal{P}(\omega)$ contains every $\Pi^1_1$ real and every $\Sigma^1_1$ real. But in fact those reals already showed up in $L_{\omega_1^{CK}+1}$, so $L_{\omega_2^{CK}}\cap\mathcal{P}(\omega)$ is much bigger than (the Boolean algebra generated by) the $\Sigma^1_1$ and $\Pi^1_1$ sets. The question is: does it reach as high as $\Delta^1_2$?
The answer, perhaps surprisingly, is no. The gulf between the first two levels of the analytical hierarchy is gigantic. The key tool here is the canonical well-ordering of $L$. For any first-order sentence $\sigma$, let $r_\sigma$ be the $\le_L$-least real coding a level of $L$ satisfying $\sigma$; this is a $\Delta^1_2$ real, and consequently all the "easily-describable" levels of the $L$-hierarchy will fall short of containing all the $\Delta^1_2$ reals. For example, let $\theta$ be the first recursively Mahlo ordinal; then letting $\sigma$ be the sentence such that the $L$-levels satisfying $\sigma$ are exactly those with recursively Mahlo index we get that $r_\sigma$ is a $\Delta^1_2$ real not in $L_\theta$ (this last bit uses the fact that if $r_\sigma$ were in $L_\theta$, then $L_\theta$ would contain a real coding $L_\theta$ itself, which via the pointwise-definability of $L_\theta$ would contradict Tarski's theorem).
Hinman's book Recursion-theoretic hierarchies contains a lot of useful information about the $\Delta^1_2$ reals and just how complicated they can be. In particular, I recommend taking a look at Corollary V.4.11, which in a precise sense says that the $\Delta^1_2$ sets can't be "built from below" in as simple a way as the hyperarithmetic sets. This old answer of mine is also relevant for getting a sense of just how powerful the second level of the analytical (also called projective) hierarchy is.