Unrigorous \begin{align*} v_7(H(n))&=\frac12\sum_{i=1}^{\lfloor\log_7(n)\rfloor}\left(7^i\left(\left\lfloor\frac n{7^i}\right\rfloor^2+\left\lfloor\frac n{7^i}\right\rfloor\right)\right)\\ &\sim\frac12\sum_{i=1}^{\lfloor\log_7(n)\rfloor}\left(7^i\left(\left(\frac n{7^i}\right)^2\right)\right)\\ &\sim\frac{n^2}2\sum_{i=1}^{\log_7(n)}\left(7^i\right)\\ &\sim\frac{n^2}2\cdot\frac{49n-49}{6n} \end{align*}and we get $\frac{49}{12}$?

@above not sure what you're doing in the last couple steps. Second last line, the summand should be $\frac{1}{7^i}$. And then this is $\sim\frac{1}{6}$. $k=\frac{1}{12}$ seems to match better with the graph.

Assume $7\vert n$ since we don't care about divisibility much as $n \to \infty$. Then we only care about the terms $7^7, 14^14, ..., n^n$ Letting $n = 7a$, we take note of how $7^2 k$ contributes $2$ powers of $7$ and $7^3 k$ contributes $3$ powers and so on... to see that $f(n) = (7 + 14 + 21 + ... + 7a) + (49 + 98 + 147 + ...) + (7^3 + 2\cdot 7^3 + ...) = 7(\frac{(a)(a+1)}{2}) + 49(\frac{\frac{a}{7}(\frac{a}{7} + 1)}{2}) + ... \approx \frac12(7a^2 + a^2 + \frac{a^2}{7} + ...) = \frac12(\frac16 n^2) \Rightarrow \boxed{\frac{1}{12}}$. $\blacksquare$