First, to clarify and to add a "yes, but..." to your statement
It is accepted that infinite order of derivatives in field theory lead to non-local effects while finite number of them local.
$$
\mathcal{L} = \frac{1}{2}\varepsilon_0 E_i(x) E_i(x)
$$
is a local Lagrangian density. To be specific: the free-field Lagrangian density of classical electrostatics. A non-local version of this could look like this:
$$
\mathcal{L} = \frac{1}{2} E_i(x) \int \varepsilon(x - x') E_i(x') \text{d}x'
$$
(where for simplicity we here assume isotropy). The corresponding material law is
$$
D_i(x) = \int \varepsilon(x - x') E_i(x') \text{d}x'\,.
$$
Now a third version could be the following Lagrangian density:
$$
\mathcal{L} = \frac{1}{2}\varepsilon_0 E_i(x) E_i(x) + \ell^2 \frac{1}{2}\varepsilon_0 \partial_jE_i(x) \partial_jE_i(x)
$$
with some parameter $\ell$ that has dimensions of length. (And in principle we could use further terms like $\partial_k\partial_jE_i \partial_k\partial_jE_i$ etc. with even higher orders of derivatives.) This is what people usually call a gradient theory, and somehow in between the local and the fully non-local theory, and some people consider this "weakly non-local".
And it can actually be obtained as an approximation to the non-local theory, because in case of a "weak" non-locality (say $\varepsilon$ being a Gaussian with very small standard deviation, or a compactly supported function with small radius of support) and $E_i$ being sufficiently smooth we can perform a moment expansion in the material law (very similar to the well-known multipole expansion):
$$
D_i(x) = \int \varepsilon(x - x') E_i(x') \text{d}x' = \int \varepsilon(x') E_i(x - x') \text{d}x' \\
\approx \int \varepsilon(x') \left[E_i(x) - x'_j \partial_jE_i(x) + \frac{1}{2} x'_jx'_k \partial_k \partial_jE_i(x) + ... \right]\text{d}x' \\
= \varepsilon_0 \left(E_i(x) - \ell^2 \Delta E_i(x) + ...\right)
$$
which gives you the material law to the third Lagrangian above, the one for the gradient theory. (For brevity I've omitted some steps here, like symmetry arguments for making some terms vanish in the expansion, defining the parameters $\ell$ and $\varepsilon_0$ as moments of $\varepsilon(x')$ etc.)
Second, to clarify the order of derivatives involved. The local Lagrangian above depends on the first derivative of the potential (because $E_i = -\partial_i \varphi$) and on the zeroth derivative of the field. There is no derivative in the material law and the field equation for the potential (Poisson's equation) contains a second-order derivative. In the gradient theory (with only the one additional term) the Lagrangian depends on the first and second derivative of the potential and on the zeroth and first derivative of the field. There is a derivative in the material law and the field equation for the potential is of fourth order. So you have to be a bit more specific as to where you have the second-order derivative you mention.
Third, to answer your question:
From the lattice theory you can actually obtain a non-local or weakly non-local theory. It's just that usually only the leading-order term is used so that you obtain the local theory (as an approximation). Because quite often this is all you need.
If you want to read a bit more on this:
https://www.sciencedirect.com/science/article/abs/pii/S0997753802012184
https://link.springer.com/chapter/10.1007/978-3-662-30257-6_37
https://rcin.org.pl/ippt/Content/84221/WA727_90438_P.262-Maugin-Nonlocal.pdf
Fourth, some bonus nitpicking on point 2:
When talking about the order of derivatives in the Lagrangian you usually take the Lagrangian among a group of equivalent Lagrangians that has the lowest orders of derivatives in it. Example: For the local electrostatic Lagrangian
$$
\mathcal{L} = \frac{1}{2}\varepsilon_0 \partial_i \varphi \partial_i \varphi
$$
and
$$
\mathcal{L} = - \frac{1}{2}\varepsilon_0 \varphi \Delta \varphi
$$
are equivalent in the sense that they yield the same Euler-Lagrange equation because they are identical up to a Null-Lagrangian (a term that can be written as a total derivative). You would use the first one, because the second one unnecessarily depends on a higher order of derivative of the potential.
