Perhaps this is easier than I thought. Fix $m \geq 1$.
The idea is the same as in the set $\{\frac 1n : n \geq 1\}$. Covering $0$ by an interval is going to take away all but finitely many points, and all we're left to do is argue about how these points are going to be covered.
Namely, let $\delta > 0$ be given as the diameter of the intervals we need to cover the set $\{\frac 1{k^m} : k \geq 1\}$ with. We must find $K(\delta)$ such that it should be impossible that both $\frac{1}{k^m}$ and $\frac 1{l^m}$ can lie together in any set of length $\delta$ for all $k,l \leq K$. In the case $m=1$, we found that $K(\delta)$ was like $\frac 1{\delta^2}$ asymptotically.
This is the same as insisting that $\inf_{k,l \leq K} |\frac{1}{k^m} - \frac 1{l^m}| > \delta$. However , the infimum on the RHS is attained precisely when $k=K-1,l=K$. In other words, we want $K(\delta)$ such that $$
\frac{1}{(K(\delta)-1)^{m}} - \frac 1{K(\delta)^m} > \delta \geq \frac{1}{K(\delta)^m} - \frac 1{(K(\delta)+1)^m} \tag{0} \label{0}
$$
Once such a $K(\delta)$ is found, it becomes clear that covering the set $\{\frac 1{k^m} : k \leq K(\delta)\}$ requires at least $K(\delta)$ boxes of diameter $\delta$.
On the other hand, we can also exhibit a covering of comparable size. To do this, consider the same $K(\delta)$ as above and cover each of the numbers $\frac 1{k^m}, k \leq K(\delta)$ with one interval of size $\delta$. Then, all the remaining elements lie inside the interval $[0,\frac{1}{(K(\delta)+1)^m}]$, which means that we require at most $\lfloor\frac{1/(K(\delta)+1)^m}{\delta}\rfloor+1$ intervals to cover this set.
In other words, we have shown that if $N(\delta)$ is the smallest number of intervals of size $\delta$ required to cover the set $\{\frac 1{k^m}, k \geq 1\}$ then $$
K(\delta) \leq N(\delta) \leq K(\delta) + \lfloor\frac{1/(K(\delta)+1)^m}{\delta}\rfloor+1 \tag{1} \label{1}
$$
All we require now are bounds on $K(\delta)$. We proceed somewhat heuristically at this point, as I plan to leave a proper, rigorous proof sketch at the end.
One easily sees by taking the common denominator that $$
\frac{1}{(K(\delta)-1)^{m}} - \frac 1{K(\delta)^m} > \delta \implies \frac{p(K(\delta))}{q(K(\delta))}>\delta
$$
where $p(x),q(x)$ are polynomials of degree $m-1,2m$ respectively in $x$. The leading coefficient of $p$ is $m$ and of $q$ is $1$, so limit heuristics tell us that we can treat this "like" $\frac{mK(\delta)^{m-1}}{K(\delta)^{2m}} > \delta$, which simplifies to $K(\delta) > \left(\frac{\delta}{m}\right)^{\frac{-1}{m+1}}$ up to a constant, but $m$ itself is a constant as $\delta \to 0$, hence we may say that $K(\delta) > \delta^{-1/(m+1)}$ up to a constant. Similarly , the other inequality tells you that $K(\delta) \approx \delta^{-1/(m+1)}$ as $\delta \to 0$, up to constants.
If we continue with the heuristics, then $$
\lfloor\frac{1/(K(\delta)+1)^m}{\delta}\rfloor \approx \frac{1/(K(\delta)+1)^m}{\delta} \approx \frac{1}{\delta \times \delta^{-m/(m+1)}} \approx \delta^{-\frac 1{m+1}}
$$
All in all, it becomes clear from \eqref{1} that $N(\delta) \approx \frac 1{\delta}^{\frac 1{m+1}}$ as $\delta \to 0$. The answer is $\frac 1{m+1}$.
Of course, I was not rigorous in parts. Therefore, I will provide below, a proof sketch that is completely rigorous.
Prove that $K(\delta) \to +\infty$ as $\delta \to 0$.
Multiply all sides of \eqref{0} by the quantity $K(\delta)^{m+1}$. This leads to $$
\frac{K(\delta)^{m+1}}{(K(\delta)-1)^m}-K(\delta) \geq \delta K(\delta)^{m+1} \geq K(\delta)-\frac{K(\delta)^{m+1}}{(K(\delta)+1)^m} \tag{2} \label{2}
$$
Observe that $$
\frac{K(\delta)^{m+1}}{(K(\delta)-1)^m}-K(\delta) = \frac{mK(\delta)^m + q(K(\delta))}{(K(\delta)-1)^m}
$$
where $q$ has degree at most $m-1$. Dividing top and bottom by $K(\delta)^m$, letting $\delta \to 0$ and recalling our first bullet point, it is clear that the LHS of \eqref{2} goes to $m$ as $\delta \to 0$. Something similar applies to the RHS of \eqref{2}, which also goes to $m$ as $\delta \to 0$.
In particular, $\delta K(\delta)^{m+1} \to m$ as $\delta \to 0$ by the squeeze theorem. Taking inverses, $\frac 1{\delta K(\delta)^{m+1}} \to \frac 1m$ as $\delta \to 0$.
Using that particular limit and a ratio comparison + squeeze argument, show that $$
\lim_{\delta \to 0} \frac{\frac 1{\delta K(\delta)^{m+1}}}{\frac 1{K(\delta)}\left(\lfloor\frac{1/(K(\delta)+1)^m}{\delta}\rfloor+1\right)} = 1
$$
Therefore, $\lim_{\delta \to 0} \frac 1{K(\delta)}\left(\lfloor\frac{1/(K(\delta)+1)^m}{\delta}\rfloor+1\right) = \frac 1m$. By the definition of limit, for some fixed $M>\frac 1m$ and all $\delta>0$ small enough, $\left(\lfloor\frac{1/(K(\delta)+1)^m}{\delta}\rfloor+1\right)<MK(\delta)$.
Plugging this into \eqref{1}, we get $K(\delta)\leq N(\delta) \leq K(\delta)(M+1)$, and taking logarithms and diving by $\log \frac 1\delta$ gives $$
\frac{\log K(\delta)}{\log\frac 1 \delta} \leq \frac{\log N(\delta)}{\log \frac 1{\delta}} \leq \frac{\log K(\delta)}{\log\frac 1 \delta}+\frac{\log(M+1)}{\log \frac 1 \delta}
$$
as $\delta K(\delta)^{m+1} \to m$, taking logarithms and rearranging gives that the right and left hand side of the above equation go to $\frac 1{m+1}$ as $\delta \to 0$. It follows that $\frac{\log N(\delta)}{\log \frac 1{\delta}} \to \frac 1{m+1}$ as $\delta \to 0$, completing the proof.
I am not fully sure about how much this will work when the structure of the set is perturbed too much. However, there is a general strategy when your set consists of a monotonic sequence $a_n$ which is decreasing to $0$. (and by translation to more general situations).
If your set $S = \{a_n : n \geq 1\}$ where $a_n$ is strictly decreasing (no point of counting the same element twice), $a_n \to 0$ as $n \to \infty$, then we can do the same thing. Given $\delta>0$, let $K(\delta)$ be the smallest positive number satisfying $$
\inf_{k,l \leq K(\delta)} |a_k - a_l| \geq \delta
$$
Then, to cover the elements $a_1,a_2,\ldots,a_{K(\delta)}$ one requires at least $K(\delta)$ elements. Now, the rest of the elements are contained in $[0,a_{K(\delta)+1}]$, therefore we can use $\lfloor\frac{a_{K(\delta)+1}}{\delta}\rfloor+1$ intervals to cover these up. We land up with the bound $$
K(\delta) \leq N(\delta) \leq K(\delta) +\lfloor\frac{a_{K(\delta)+1}}{\delta}\rfloor+1
$$
Now, suppose it happens that $\lfloor\frac{a_{K(\delta)+1}}{\delta}\rfloor+1 \leq MK(\delta)$ for some constant $M>0$ and small enough $\delta>0$. We can easily see that the box-counting dimension equals $\lim_{\delta \to 0}\frac{\log K(\delta)}{\log \frac 1{\delta}}$ in this case, whenever that limit exists.
What happens if this does not occur? That can only happen when the $a_n$ continue to be really spaced "far apart" as $n \to \infty$, so that the brute force covering by adjacent intervals of size $\delta$ becomes inefficient and we have to account for gaps in the $a_n$ which could be abnormally large. This is possible in a fractal-like situation, but in a situation where there is a lot of monotonicity (for example, I think insisting that $a_{n} - a_{n+1}$ is a decreasing sequence as well should be good enough) it shouldn't occur.
I am not sure what can be said about infinitely many unions of sets and the box-counting dimension, but in this particular scenario it is clear that the box-dimension should be computable in a large number of situations.
