As mentioned in answers and comments, the definition given for a boundary point is not the correct one. A possible one should read:
A point $p \in M$ is a boundary point if it is such that $\pi_n(\phi(p))=\phi_n(p)=0$ for some chart $\phi:U \to \mathbb{R}^n_{+}$.
Then, I think, you would be willing to prove that if this is true for one chart $\phi: U \to \mathbb{R}^n_+$ as per the definition above, then it is true for any such chart. As one other answer suggests and references, this is not a big deal in the smooth setting.
This is true in the topological setting too, but due to a more involved reason: namely, the invariance of domain theorem. I don't know a reference which manages to avoid this.
Now, moving forward: Indeed, let $\psi: V \to \mathbb{R}^n_+$ be another chart. What we must prove then is that $\psi_n(p)=0$.
Instead of doing that, let's prove that if $\phi_n(p)>0$, then $\psi_n(p)>0$. It is easy to see (exchange places of $\phi$ and $\psi$) that this will prove that $\phi_n(p)>0$ if and only if $\psi_n(p)>0$, thus concluding that $\phi_n(p)=0$ if and only if $\psi_n(p)=0$.
Thus, suppose $\phi_n(p)>0$. We have that $\psi \circ \phi^{-1}$ is a continuous injection from $U \cap\mathbb{R}^n_+$ to $\mathbb{R}^n_+$, where $U$ is an open subset of $\mathbb{R}^n$. Restricting, $\psi \circ \phi^{-1}|_{U \cap \mathbb{R}^n_{>0}}: U \cap \mathbb{R}^n_{>0} \to \mathbb{R}^n$ is then injective and continuous. By invariance of domain, we know that such an image is open in $\mathbb{R}^n$. But this image is inside $\mathbb{R}^n_+$, therefore it can't intersect the set $\{x \mid x_n=0\}$. Since $\psi(\phi^{-1}(\phi(p)))=\psi(p)$ is in such image, it follows that it can't be such that $\psi_n(p)=0$.
As a sidenote which may also be of interest, a similar use of invariance of domain also yields that if $p$ is a boundary point, then there is no chart $\phi: U \to \mathbb{R}^n$ around $p$ (which is a homeomorphism with $\mathbb{R}^n$).
It is not true. Take any smooth manifold (which is not a sphere) $X$. Consider the cone C(X). Embed it into some higher $\mathbb R^n$. And now consider $N= C(X)- {\ cone \ point}$. Then $\bar{N}= C(X)$ which is not a manifold (WHY?).
Using this similar idea, you can construct many more examples whose closure's boundary is a smooth submanifold but not the closure one.
I don't think it is even true in low dimension i.e 3. For example consider an open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by annulus, and their closure is no longer a surface.
EDIT Another explicit example: Consider the lens space $L(3,1)$. It can be build by gluing a $D^2$ to $S^1$ by a degree three map form $\partial D^2 \to S^1$. Then glue on an B^3. It is not hard to see that the interior of D^2 is a submanifold of $L(3,1)$ with one end. It’s closer is not a submanifold.
Some geometric curvature bound can ensure it to be an embedded manifold.
Best Answer
(See around page 189 of J.M. Lee's Introduction to Smooth Manifolds. I assume you are worried about smooth manifolds, since you tagged differential-topology.)
Recall that the tangent bundle of a smooth manifold with boundary is defined the same way as the tangent bundle of one without boundary in the interior, and on a boundary point $p\in\partial M$ we take $T_pM$ to be the span of $\{\partial_1, \ldots \partial_n\}$ ($\dim M = n$) in the boundary chart. So the tangent space at any point on a smooth $n$-manifold with boundary is an $n$-dimensional vector space.
Thus we can define immersions between manifolds with boundary: a smooth map $F:M\to N$ where $M, N$ are manifolds (possibly with boundary) is said to be an immersion if the differential $dF|_p:T_pM \to T_{F(p)}N$ is injective for every $p\in M$. Then an immersed submanifold of a manifold with boundary is simply $S\subset M$ such that $S$ can be given a smooth structure (possibly with boundary) such that the inclusion map $\iota:S\to M$ is an immersion. Similarly you can define embedded submanifolds.
In particular, for a smooth manifold with boundary $M$, the boundary $\partial M$ is a smooth submanifold of $M$. Similarly, under Lee's definition, your example is also a submanifold.
(Slightly off-topic:) Though honestly, instead of asking whether Blah is a Foobar, which really strikes me as a job for taxonomists, you should be thinking about the practical implications of the definitions. For example, while it is true still that for any embedded submanifold of a manifold with boundary that you still can find an open neighborhood of it that deformation retracts to the submanifold, it is not true that we can have the "tubular neighborhood theorem". (That is, it is not true that a sufficiently small open neighborhood of the embedded submanifold is diffeomorphic to its normal bundle; any neighborhood of $\partial M$ is a manifold with boundary, whereas the normal bundle of $\partial M$ is a manifold without boundary.)