I think the issue here is that you need to keep a consistent level of approximation in your "small angle approximation." By small angles, we typically mean $\theta_1$ and $\theta_2$ are both of order $\epsilon$, where $\epsilon \ll 1$. Then the question is - to what order in $\epsilon$ do you want to write down the equations of motion?
When you neglect the term $\frac{3}{2} \dot\theta_1 \dot \theta_2 (\theta_1 - \theta_2)^2$ in the Lagrangian, you are saying that terms of size $\epsilon^4$ are small compared to terms like $\dot \theta_1^2$, which is of size $\epsilon^2$. In the equation of motion, you get terms that are $\dot \theta_1 \dot \theta_2 (\theta_1 - \theta_2)$, which are of size $\epsilon^3$, compared to $\theta_1$, which is size $\epsilon$.
So neglecting the additional term in the Lagrangian gets you the same equation of motion as keeping the whole Lagrangian, and then dropping terms that are of size $\epsilon^3$.
This kind of argument is a little handwavy, and (in principle) could blow up if the time derivatives of $\theta_{1,2}$ were large - at some point, you might want to check out some books on perturbation theory in a more formal sense.
i will try this one.
A Hamiltonian system is (fully) integrable, which means there are $n$ ($n=$ number of dimensions) independent integrals of motion (note that completely integrable hamiltonian systems are very rare, almost all hamiltonian systems are not completely integrable).
What this states in essence (and intuitively) is that the hamiltonian system of dimension $n$ can be decomposed into a cartesian product of a set of $n$ independent sub-systems (e.g in action-angle representation) which are minimally coupled to each other.
This de-composition into a cartesian product of $n$ independent systems (each of which has bounded energy as the whole system has bounded energy), means topologically is the $n$-dimensional torus $S^1 \times S^1 \times ... \times S^1$ ($n$ factors) which is compact (bounded system is topologically compact).
note $S^1$, literaly means topological circle or topological $1$-dimensional sphere. What it means, is that it represents (since this is topology and not geometry) a compact, bounded 1-dimensional space (1-parameter space). So a hamiltonian system with $n$ independent parameters (integrable) is (should be, locally) topologicaly the cartesian product of $n$ (abstract) $S^1$ spaces ($1$ for each parameter/dimension)
Each $S^1$ space represents a simple harmonic oscilator (a simple periodic system, or in other words a system moving on a circle, see the connection with $S^1$ spaces).
When a (completely) integrable hamiltonian system is de-composed into $n$ indepenent sub-systems, in essence this means that (locally, at every neighborhood of a point of the system phase-space) it can be linearised and represented as a stack of (independent) harmonic oscilators (stacks of $S^1$ spaces). This is the basic theorem of Liouville-Arnold on hamiltonian dynamics
For a simple example of a 3-dimensional (actually 2-dimensional, since the configuration space is the surface of a sphere) hamiltonian system which is completely integrable, see the spherical pendulum and analysis thereof
The spherical pendulum is 2-dimensional system (thus the phase-space is 4-dimesnional) and has a second integral of motion the moment about the vertical axis.
(a link on a more advanced analysis on the dynamics of pendula).
In other words the whole is just the sum of its parts.
What would be the hamiltonian space of a (for example $2$-dimensional) system which the dimensions are not independent (not-integrable).
This means the dimensions are correlated and cannot be de-composed into independent sub-systems (i.e a $2$-dimensional torus $S^1 \times S^1$), so topologically it is a $2$-dimensional sphere ($S^2$).
In a $2$-dimensional sphere the $2$ dimensions are correlated and cannot be made flat (i.e cannot be linearised and mapped into a flat space of same dimension, unlike a $2$-dimensional torus, in other words has what is refered as intrinsic curvature).
Elaborating a little on this.
Of course, if one sees the 2-dimensional torus as a 3D object (in effect this means embedded in a flat 3D euclidean space), it has curvature. This is refered as "external" curvature stemming from the embedding into a 3D space. But if one sees the 2-dim torus as a 2-dimensional surface on its own, it has no (zero) curvature. This is refered to as (intrinsic) curvature (in the riemannian sense).
If one takes the 2-dim torus and cuts it and unfold it, one gets the 2-dimensional cylinder
. If further one cuts the 2-dim cylinder and unfold it, one gets a 2-dimensional flat surface. This means the (intrinsic) curvature of the 2-dim torus is zero and can be mapped into a flat space of the same dimension.
For the 2-dim sphere, this is not possible. There is no way it can be cut and mapped into a flat surface of the same dimension. It has (intrinsic) curvature non-zero and this is also a measure away from flatness (and also a measure of dimension correlation). One example is maps of earth (2-dimensional spherical surface) on a flat paper, one can see that the map contains distortions, since there is no mapping of a sphere into a flat surface.
On the other hand if one takes a flat 2-dim surface and makes one boundary periodic, one gets a 2-dim cylinder, if further makes the other boundary also periodic, one gets the 2-dim torus.
In general the conditions under which any given hamiltonian system is (completely) integrable is a very difficult problem.
Still another way to see this is an analogy with probability spaces. Consider 2 event spaces of 2 physical systems consisting of 2 parameters (lets say 2 coins) $\Omega_{12}$ and $\Omega_{AB}$.
When the system is integrable (i.e the parameters are independent, meaning $P(1|2) = P(1)$) then the event space $\Omega_{12}$ is the cartesian product of each sub-space $\Omega_1 \times \Omega_2$. And each outcome of the total system is just the product of the probabilities of each sub-system.
Now consider a second system where the coins are correlated, meaning $P(A|B) \ne P(A)$.
This space $\Omega_{AB}$ cannot be de-composed into 2 independent sub-spaces $\Omega_A$, $\Omega_B$ as their cartesian product since the sub-spaces are not independent. This corresponds to a non-integrable Hamiltonian system (and a topological $2$-d sphere).
The analog of statistical independence in probability event spaces for hamiltonian systems is exactly the existence and functional (more correctly poisson) independence of the appropriate number of integrals of motion (complete integrability).
In other words for a non-integrable system the whole is more than the sum of its parts.
Hope this is useful to you
PS. You might also want to check: Holonomic System, Non-holonomic System, Integrable System
Best Answer
To start things off I'd say that noting the $L_z$ component is conserved seems to mean pretty much nothing, since you're considering the motion as restricted to the $\mathcal{X}\mathcal{Y}$ plane. If you had assumed the motion along the $\mathcal{Z}$ axis to be possible, then we'd be talking about the spherical double pendulum instead of the planar one (which is the case, since the lagrangian has two degrees of freedom).
The energy will be conserved because the system is autonomous (time-independent). Notice also that an integrable autonomous system with $n$ freedoms has $n$ conserved quantities, one of them being the energy. So, since our system has two degrees of freedom, there is one constant of motion missing for it to be integrable. Differentiating $L$ with respect to $\dot{\theta}_1$ and $\dot{\theta}_2$ will give us the canonical momenta of the system, but notice that the derivatives of $L$ with respect to $\theta_1$ and $\theta_2$ are not zero. Thus, the canonical momenta are not conserved quantities. Also, the total energy of the system doesn't factor as a sum of individual energies, since the Lagrangian has mixed terms in it. You cannot extract any other quantity from the Lagrangian which should be conserved, since mechanics talks about the conservation of momenta and energy (and sometimes their projections). Since there are not as much conserved quantities as there are degrees of freedom, the system is non-integrable.
P.S.: In one dimension we can clearly see that (using simple examples as the harmonic oscillator or the simple pendulum) some systems do not conserve their momenta. Even though, if they're autonomous then they're integrable, because they have one degree of freedom and one conserved quantity: energy.
EDIT.: Since the question is directly oriented to "is it possible to obtain, from the Lagrangian, an answer about integrability?", then (as suggested in the comments) I'll sketch something on Noether's theorem. The theorem relates Lie groups to invariant quantities, so it's just one more way to find conserved quantities. It basically says that if you find a transformation that leaves the Lagrangian invariant, then there's a conserved quantity associated to that transformation. As an example, when the transformation reduces to a translation, invariance of the Lagrangian implies momentum conservation; much the same way, if the transformation is a rotation, then invariance implies the conservation of angular momentum along the axis of rotation. So this is basically a way of using symmetries of the Lagrangian to obtain conservation laws (knowing Noether's theorem content well is very important to a clear understanding of many concepts of Quantum Mechanics and Quantum Field Theory). I'm not being quantitative because proving the double pendulum's Lagrangian is not an invariant is tedious, even thought it should be somehow obvious by just looking at it.