What general properties are known of the function defined as follows: Let $H:[1,\infty[\to\mathbb{R}$ be given by $H(t):=\mathcal{H}^t([0,1],\|\cdot\|^\frac{1}{t})$?
Given the difficulty of computing the Hausdorff measure of the Koch snowflake I don't expect explicit values, just general non trivial results.
For any metric space $(X,d)$, any $s>0$ and any $t\ge 1$ we have $$ \mathcal H^{ts}(X, d^{1/t}) = \mathcal H^s(X, d) $$
Indeed, let $(E_k)$ be a cover of $X$ by sets of small diameter that almost achieves $\mathcal H^s(X, d)$, that is $$ \sum_k (\operatorname{diam}E_k)^s \le \mathcal H^s(X, d) + \epsilon $$
In the snowflaked version of $X$ (distinguished by subscript $t$), $\operatorname{diam}_tE_k = (\operatorname{diam} E_k)^{1/t}$. Hence $$ \sum_k (\operatorname{diam}_tE_k)^{ts} = \sum_k (\operatorname{diam}E_k)^s $$ which proves $\mathcal H^{ts}(X, d^{1/t}) \le \mathcal H^s(X, d)$. The reverse inequality follows in the same fashion.
Summary
Abstract "snowflaking", by raising the metric to a power, is easy. The difficulty with von Koch curve is that it's not exactly the same as raising the metric on an interval to a power (it's only bi-Lipschitz equivalent). Concrete realizations of snowflakes are more difficult to handle than abstract ones: there are all these bends in various directions, and it's not obvious what the optimal covers should look like.