Minkowski Dimension of Special Cantor Set

I think you're making this a bit harder than it needs to be. You don't need an exact formula for $\text{vol}(C_{\delta_n})$, just upper and lower bounds. Furthermore, these can be quite crude - within a constant multiple.

To make this precise and to set notation, define an $\varepsilon$-mesh interval to be one of the form $(j\varepsilon,(j+1)\varepsilon)$ and define $N_{\varepsilon}(C')$ to be the number of $\varepsilon$-mesh intervals that intersect $C'$. The upper and lower box-counting dimensions of $C'$ are then $\limsup$ and $\liminf$ as $\varepsilon\to0^{-}$ of $$\frac{\log(N_{\varepsilon}(C'))}{\log(1/\varepsilon)}.$$ Note that I've chosen to use open intervals, since that will ease some further computations. If I had used closed intervals to obtain a related quantity, say $\overline{N}_{\varepsilon}$, then we could increase the value by as much as a factor of $3$. Ultimately, this doesn't affect the dimension computation since $$\frac{\log(N_{\varepsilon}(C'))}{\log(1/\varepsilon)}\leq \frac{\log(\overline{N}_{\varepsilon}(C'))}{\log(1/\varepsilon)} \leq \frac{\log(N_{\varepsilon}(C'))+\log{(3)}}{\log(1/\varepsilon)}$$ and that extra $\log(3)$ has no bearing on the limit. Also note that there is an easy relationship between these quantities and volume of the $\varepsilon$ fattening, as Tao shows later on that same page.

We'll now show that the lower box-counting dimension of $C'$ is $1/2$; the other computation is similar. To do so, it is sufficient to show that $$\liminf_{k\to\infty} \frac{\log(N_{\varepsilon_k}(C'))}{\log(1/\varepsilon_k)}\leq \frac{1}{2}$$ for $\varepsilon_k = 4^{-((2k+1)!-1)}$, as you've already mentioned that we know the lower box-counting dimension is at least $1/2$. To do this, we need an upper bound on $N_{\varepsilon_k}(C')$.

First, an easy upper bound on $N_{4^{-(2k)!}}(C')$ is $4^{(2k)!}$. Next, for each increment of $i$ over the span $$(2k)!\leq i < (2k+1)!,$$ $N_{4^{-i}}(C')$ increases by the factor $2$. Furthermore, $$(2k+1)!-(2k)! = (2k)!((2k+1)-1) = 2k(2k)!.$$ Thus, $$N_{4^{-((2k+1)!-1)}}(C') \leq 4^{(2k)!} 2^{2k(2k)!} = 4^{(2k)!(k+1)}$$ and $$\frac{\log(N_{4^{-((2k+1)!-1)}}(C'))}{\log\left(4^{((2k+1)!-1)}\right)} \leq \frac{(2k)!(k+1)\log(4)}{(2k+1)!\log(4) - \log(4)} \to \frac{1}{2}. $$