If $M$ is a topological $n$-manifold, then I know that there is a metric $\rho$ (compatible with the topology) such that the Hausdorff dimension of $(M,\rho)$ is $n$ (see this question). Is this also true for the Minkowski (upper box counting) dimension? I wouldn't be surprised if not, as it seems to behave much more volatile than Hausdorff dimension. In that case, is there always a metric such that the Minkowski dimension is at least finite?
If this is not easy to answer, are there some additional assumptions on $M$ that change this (differentiable, smooth etc.)? I know that Falconer mentions that the Minkowski dimension is always $n$ if $M$ is a smooth submanifold of some $\mathbb{R}^m$, for example.
It is a theorem due to Edward Szpilrajn (Marczewski),
E. Szpilrajn, La dimension et la mesure", Fund. Math. 28 (1937) 81–89.
that every separable metrizable topological space of covering dimension $n$ is homeomorphic to a subset $E$ of ${\mathbb R}^{2n+1}$ whose Hausdorff dimension is $n$. You can find an English-language proof of this result in the book
Hurewicz, Witold; Wallman, Henry, Dimension theory, Princeton: Princeton University Press. 165 p. (1948). ZBL0036.12501.
(I will add a precise reference later: I do not have access to the book right now.)
At the very least, it follows that the Minkowski dimension of $E$ is at most $2n+1$, in particular, finite. (There is a good chance that the proof shows more, namely that it is $=n$.) In particular, every $n$-dimensional manifold can be metrized by a metric of Minkowski dimension $\le 2n+1$.