No, there are not. For instance, there exists a sequence of $\omega_1$ functions $f_\alpha:\mathbb{N}\to\mathbb{N}$ (for $\alpha<\omega_1$) such that if $\alpha<\beta$ then $f_\alpha(n)<f_\beta(n)$ for all sufficiently large $n$. (Proof sketch: given any countable set of functions, you can build a function that is eventually larger than each of them by a diagonal argument. So you can construct a sequence of length $\omega_1$ by transfinite recursion, choosing each new term of the sequence to be eventually larger than the previous ones.)
Now let $M$ be a nonstandard model of $\text{Tr}(L_\text{full})$ and let $n\in M$ be any nonstandard element. Then the sequence $(f_\alpha(n))$ must be strictly increasing, so this gives $\aleph_1$ different elements of $M$. Thus $M$ must be uncountable.
Here's another argument that is perhaps more elementary and gives a stronger result (thanks to Alex Kruckman for suggesting a variant of this in the comments). For each real number $r>0$, consider the function $f_r:\mathbb{N}\to\mathbb{N}$ defined by $f_r(n)=\lfloor rn\rfloor$. Note that the ratio $f_r(n)/n$ approaches $r$ as $n\to\infty$. It follows that if $n$ is a nonstandard element of a model, the elements $f_r(n)$ must all be distinct, since we can recover $r$ as the Dedekind cut in $\mathbb{Q}$ defined by comparing multiples of $n$ and multiples of $f_r(n)$. So, the model must have at least $2^{\aleph_0}$ elements.
Finally, let me discuss a generalization. As your question really only involves $\mathbb{N}$ as a set, it is natural to ask the same question with $\mathbb{N}$ replaced by any infinite set $X$: what is the smallest possible cardinality of a nonstandard model of the theory of $X$ with respect to its full language (I will call this the "full theory of $X$")? Note first that any countable ultrapower of $X$ will be a nonstandard model of the full theory of $X$ of cardinality $|X|^{\aleph_0}$. (In general, this bound is better than the bound $2^{|X|}$ given by Löwenheim-Skolem, and in many cases is equal to just $|X|$!)
However, this bound is not sharp in general. For instance, suppose $\kappa$ is a measurable cardinal and let $\lambda>\kappa$ be strong limit cardinal of cofinality $\omega$ (actually, all we need from "strong limit" is that $\theta^{\kappa}\leq\lambda$ for all $\theta<\lambda$). Let $U$ be a countably complete ultrafilter on $\kappa$ and let $M$ be the ultrapower of $\lambda$ with respect to $U$. Since $\lambda>\kappa$, $M$ is a nonstandard model of the full theory of $\lambda$. However, since $\lambda$ has cofinality $\omega$ and $U$ is countably complete, every element of $M$ is represented by a bounded function $\kappa\to\lambda$. The number of such functions is $\lambda\cdot\sup_{\theta<\lambda}\theta^\kappa=\lambda$, so $|M|=\lambda$. In particular, $|M|<\lambda^{\aleph_0}$ since $\lambda$ has cofinality $\omega$.
Certainly the good money is on the (current) nonexistence of such an example. But that's boring. Here are a couple positive observations, although - in my opinion - each falls short of an actual positive answer, and indeed I don't know of anything I would call a "natural" nonstandard model of $\mathsf{PA}$.
First, here's a basic model-theoretic observation. Say that a structure $\mathfrak{M}$ is pointwise-definable iff every element of $\mathfrak{M}$ is (parameter-freely-)definable in $\mathfrak{M}$. Some structures are pointwise-definable (e.g. $(\mathbb{N};+,\times)$ or $(\mathbb{Q};+,\times)$) while others aren't (e.g. $(\mathbb{R};+,\times)$ or $(\mathbb{Q};+)$). Similarly, some but not all theories have pointwise-definable models, and some theories have pointwise-definable models even though we might reasonably expect them not to (such as $\mathsf{ZFC}$). Of course, whenever a complete theory does have a pointwise-definable model it has exactly one up to isomorphism (just "match up the definitions" and check that everything works).
Now for a structure $\mathfrak{A}$, let $\mathfrak{A}^{\mathsf{def}}$ be the substructure (allowing emptiness) of $\mathfrak{A}$ consisting of all definable elements (ignoring the issue that the latter might be empty). In general $\mathfrak{A}^{\mathsf{def}}$ might be very unlike $\mathfrak{A}$; however, it turns out that models of $\mathsf{PA}$ behave very nicely in this regard: we always have $\mathfrak{A}^{\mathsf{def}}\preccurlyeq\mathfrak{A}$ when $\mathfrak{A}\models\mathsf{PA}$. This is because we have full induction in $\mathsf{PA}$: think about the Tarski-Vaught test and note that the least number satisfying a formula is always definable.
Consequently every completion of $\mathsf{PA}$ has a pointwise-definable model. By the uniqueness observation above this means that arguably every natural completion of $\mathsf{PA}$ (other than true arithmetic, $\mathsf{TA}$, itself) gives rise to a natural nonstandard model of $\mathsf{PA}$. Of course we still need to come up with a "natural" completion of $\mathsf{PA}$ other than $\mathsf{TA}$, and I don't really think it's clear how to do that, but I do think this is still a positive-"flavored" result in some sense.
What about nonstandard models of $\mathsf{TA}$? The unique pointwise-definable model of $\mathsf{TA}$ is obviously just $\mathbb{N}$ itself, so the above doesn't help. Here we can turn to ultrapower constructions, and - despite the obvious obstacles - there are still a couple of positive things to be said.
First, assuming $\mathsf{CH}$ we have that any two nonprincipal ultrafilters on $\omega$ yield isomorphic ultrapowers of $\mathbb{N}$ (or indeed any other countable structure - see here). So "the nontrivial $\omega$-fold ultrapower of $\mathbb{N}$" is actually a proper description of a nonstandard model of arithmetic, and the non-canonical nature of the ultrafilter used turns out to disappear.
In $\mathsf{ZFC}$ alone a more involved construction appears: not a single ultrapower, but a way of combining all possible ultrapowers in a sense. In broad terms, Kanovei and Shelah produced a way to construct, unambiguously, a hyperreal field which works just assuming $\mathsf{ZFC}$. The natural numbers part of this field then is a nonstandard model of $\mathsf{TA}$. I don't personally consider this a "natural" nonstandard model, but it is certainly highly interesting and canonical in some sense.
OK, one last shot:
Solving all our problems by getting rid of the axiom of choice
(Wait, what in the whatting what?!)
Working in $\mathsf{ZF}$ alone, it is consistent that there are infinite cardinalities (which noun incidentally becomes a bit more subtle than in choice-land) which behave a bit like natural numbers. Specifically, Sageev showed the following:
Suppose that $\mathsf{ZFC}$ + "There is an inaccessible cardinal" is consistent. Then there is a model of $\mathsf{ZF}$ + "There is a Dedekind-finite infinite set" in which the Dedekind-finite cardinalities are linearly ordered by injectability.
This turns out to be highly relevant to our question due to earlier (and more elementary but harder to find) work by Ellentuck:
In any model of $\mathsf{ZF}$ where the Dedekind-finite cardinalities are linearly ordered by injectability, they in fact form a nonstandard model of $\mathsf{TA}$ (with "$+$" and "$\times$" coming from disjoint union and Cartesian product as expected).
Put together these results show that certain models of $\mathsf{ZF}$ come with in my opinion highly natural and surprising canonical nonstandard models of true arithmetic. Of course this requires choice to fail extremely badly and much about these models is still unclear (see 1, 2), so I still don't think this constitutes a positive answer to your question, but still - pretty dang cool, no?
Best Answer
For every natural $n$, $\phi_n$ is a sentence, where $\phi_0$ is $\forall x\,(x=x)$ and $\phi_{n+1}$ is $(\phi_n\land\phi_n)$. By recursion, there is a sentence in the theory that codes this claim and so, for any model, for any $n$ that, from the point of view of the model, is a natural number, there is an object of the model that the model interprets as the sentence $\phi_n$. This holds even if $n$ is nonstandard.
Of course, if $n$ is nonstandard, this object $\phi_n$ is not really a formula, but the model cannot see that.