While studying locally compact Hausdorff spaces, I found the following result:
Let $X$ be a second countable LCH space. Then there exists a complete metric $d : X \times X \to \mathbb{R}$ for the topology of $X$ such that every open ball $B_d(x,\epsilon)$ is precompact.
This corresponds to the result in $\mathbb{R}^m$ that every Euclidean open ball is precompact.
In particular, since a manifold $M$ is a second countable LCH space (provided we define manifolds as being Hausdorff, second countable, locally Euclidean spaces), the above result holds to yield a "nice" metric $d : M\times M \to \mathbb{R}$. But even though this $d$ is topologically nice, its construction is agnostic of the local Euclidean structure on $M$. For example, in $\mathbb{R}^m$, open balls are actually chart domains, but this need not hold for our $d$ on $M$.
Nevertheless, there are nontrivial examples of manifolds that admit complete metrics whose open balls (of sufficiently small radius) are all chart domains. I've worked with the following "arc metric" on the circle $S^1$: $$ d_{arc}(t,s) = \min\{|t-s|, 1 - |t-s|\}, $$ where we treat $S^1$ as $[0,1)$ with a non-standard topology. Then $B_{d_{arc}}(t, \epsilon)$ is actually a smooth chart domain for $S^1$ if $\epsilon < 1$. And for any $\epsilon$, $B_{d_{arc}}(t,\epsilon)$ is at least a $1$-submanifold of $S^1$. I believe this example can be extended to $S^n$, more generally.
My question: Given a manifold $M$, does there exist a complete metric $d : M\times M\to \mathbb{R}$ for the topology of $M$ such that for any $p\in M$ (sufficiently small) $\epsilon > 0$, the open $\epsilon$-ball at $p$ is a precompact chart domain for $M$? If $M$ is further taken to be $C^k$, can we force the open $\epsilon$-ball to be $C^k$ also? Separately, can we say (the closure of) $B_d(p,\epsilon)$ is at least a submanifold (with boundary)?
Let me assume that $M$ is a smooth, connected manifold. Then, your question has a positive answer using Riemannian geometry. For the details, I refer to any textbook on the subject, so let me merely outline this positive answer. The manifold $M$ possesses a Riemannian metric, i.e. a smooth choice of inner product on every tangent space. This means roughly two things:
a) The length of a smooth path $\gamma\colon I=[0,1]\rightarrow M$ can be defined by integrating the norm of its derivative. Then, defining $d(p,q)$ to be the infimum of lengths of all (piece-wise) smooth paths from $p$ to $q$ defines a metric on $M$ inducing the original topology.
b) There is a notion of (locally) length-minimizing curves called geodesics. For every point $p\in M$, there is an exponential map $\exp_p\colon O\rightarrow M$, where $O\subseteq T_pM$ is some open, star-shaped subset with respect to the origin. This maps a tangent vector $v$ to the point $\gamma_v(1)$, where $\gamma_v$ is the unique (by ODE theory) geodesic starting at $p$ with velocity $v$. The exponential map is smooth and its derivative at $0$ is the identity, so it is a local diffeomorphism. For sufficiently small $\varepsilon>0$, the open ball $U_{\varepsilon}(0_p)\subseteq T_pM$ thus gets mapped diffeomorphically to an open neighborhood $p\in U_{\varepsilon}$ in $M$.
For a given $q=\exp_p(v)\in U_{\varepsilon}$ in this neighborhood, it is elementary to show that the curve $t\mapsto\exp_p(tv)$ is a geodesic from $p$ to $q$ and has length $\lVert v\rVert<\varepsilon$, so $d(p,q)<\varepsilon$. (In fact, the metric in exponential coordinates is described by the Gauss lemma, which implies that $d(p,q)=\lVert v\rVert$ exactly.) Thus, $U_{\varepsilon}$ is contained in the open $\varepsilon$-ball about $p$. If, conversely, $q\not\in U_{\varepsilon}$, let $\varepsilon^{\prime}<\varepsilon$ be arbitrary. Any path from $p$ to $q$ has to cross the "geodesic sphere" $\exp_p(\partial B_{\varepsilon^{\prime}}(0_p))$ since it disconnects $M$ with $p$ and $q$ lying in different components. The Gauss lemma however already implies that the distance between $p$ and this point of crossing is already $\varepsilon^{\prime}$, so $d(p,q)>\varepsilon^{\prime}$. Since $\varepsilon^{\prime}$ was arbitrary, we conclude $d(p,q)\ge\varepsilon$. Thus, $U_{\varepsilon}$ is precisely the open $\varepsilon$-ball about $p$. You can deduce analogous statements about the closed balls from this.
In conclusion, if you equip $M$ with the metric $d$ induced by a Riemannian metric, then, for any $p\in M$, the open/closed metric balls about $p$ of sufficiently small radius are diffeomorphic to open/closed Euclidean balls. In fact, with a more careful analysis, you can show that the "sufficiently small radius" can be chosen independently of $p$, as long as $M$ is compact.
Edit: I neglected that the question wanted a complete metric. Here's one way to get a complete such metric: A weak Whitney theorem implies that there is a closed smooth embedding $M\hookrightarrow\mathbb{R}^N$ for some $N\gg0$. Equip $M$ with the pullback metric of the standard Riemannian metric on $\mathbb{R}^N$. The resulting induced metric $d$ on $M$ is generally not the restriction of the Euclidean metric to $M$, yet it is clearly bounded below by it. This suffices to imply that any closed, bounded subset of $M$ is also a closed, bounded subset of $\mathbb{R}^N$, hence compact by the Heine-Borel theorem. This in turn implies that $(M,d)$ is complete.