First off, the solid Koch Snowflake is, in fact, self-similar; it consists of seven copies of itself - six of which, shown in gray in figure below, are scaled by the factor $1/3$ and one of which, shown in red in the figure below, is scaled by the factor $1/\sqrt{3}$.
The formula that you mention,
$$
\text{dimension} = \frac{\log(\text{number of self-similar pieces})}{\log(\text{magnification factor})},
$$
works only for simpler sets, where all the pieces have the same scaling factor. More generally, a self-similar set can consist of $N$ copies of itself scaled by the factors $\{r_1,r_2,\ldots,r_N\}$. In this case, the similarity dimension of the set is defined to be the unique value of $s>0$ such that
$$
r_1^s + r_2^s + \cdots + r_N^s = 1.
$$
Note that if $r_1=r_2=\cdots=r_N = r$, then the equation simplifies to $N r^s=1$. In this case, you can solve for $s$ to get the simpler formula.
For the solid Koch curve, we expect the dimension to be 2. In fact, if we set $s=2$ and use the scaling factors for the solid Koch flake, we get.
$$
6 (1/3)^2 + (1/\sqrt{3})^2 = 2/3+1/3 = 1,
$$
as expected.
I want to post the solution, since someone might need it.
Proof. by ordinary induction. Let Induction hypothesis $P(n)$ be $$a_n = a_0\left(\frac{8}{5} - \frac{3}{5}\cdot \left(\frac{4}{9}\right)^n\right).$$
Base case $(n=0):$ $a_0=a_0\left(\frac{8}{5} - \frac{3}{5}\cdot \left(\frac{4}{9}\right)^0\right) = a_0.$ holds.
Inductive step: Assume $P(n)$ holds for some $n \geq 0$. Show that $P(n) \Rightarrow P(n+1).$ We need to show $$a_{n+1} = a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^{n+1}\right).$$ We can write $$a_{n+1} = a_n + e_n t_{n+1}$$ where $e_n = 3\cdot4^n$ and $t_{n+1} = \frac{a_0}{9^{n+1}}.$ Replacing $e_n$ and $t_{n+1}$ in the equation of $a_{n+1}$ gives $$a_{n+1} = a_n + 3\cdot4^n\cdot\left(\frac{a_0}{9^{n+1}}\right)=$$ $$a_n + 3\cdot\left(\frac{1}{9}\right)\cdot a_0\cdot\left(\frac{4^n}{9^{n}}\right)=$$ $$a_n + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ Since P(n) holds, we can replace $a_n$ with $a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^n\right).$ $$a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^n\right) + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot \left(\frac{3}{5}\right)\cdot\left(\frac{4}{9}\right)^n + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \left(\frac{3}{5} - \frac{1}{3}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \cdot \left(\frac{4}{15}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \cdot \left(\frac{4}{9} \cdot \frac{3}{5}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^{n+1} \cdot \frac{3}{5}=$$ $$a_0\cdot \left(\frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^{n+1}\right)$$ which proves $P(n+1).$
We can conclude, by induction principle, $\forall n \geq 0: P(n)$.
Best Answer
The 2007 paper you linked in a comment, Bounds of the Hausdorff measure of the Koch curve, contains information which is the most recent I could find. In it, and the related Bounds of Hausdorff measure of the Sierpinski gasket, both by Baoguo Jia, an approach to estimating Hausdorff measures of self-similar sets satisfying the open set condition (citing some books by Falconer on fractal geometry for some facts about the Hausdorff measure for the proof that this method works) is outlined. However, the method is not very effective (it is not easy to calculate good approximations with certainty).
In "Bounds of the Hausdorff measure of the Koch curve", the author proves that the $s$-dimensional (where $s=\log4/\log3$) Hausdorff measure of the Koch curve (base length 1) is bounded below by $$\left(2 \left(\frac{2 \sqrt{3}}{9}\right)^s\right) \exp\left(-\frac{12 s\sqrt{3}}{9}\right)=2\text{^}\left(-2-\frac{8}{\sqrt{3} \log (3)}+s\right)\approx0.0325239$$and bounded above by $$2 \left(\frac{2 \sqrt{3}}{9}\right)^s=2^{s-2}\approx0.599512\text.$$
At the end of the paper, they conjecture tighter bounds. Assuming their $6/81$ was meant to be ${\sqrt{876}}/{81}$ (the context makes this a reasonable typo), they conjecture a lower bound of $$\left(\frac{1}{122} 4^4 \left(\frac{\sqrt{876}}{81}\right)^s\right) \exp\left(-\frac{12 s\sqrt{3} }{3^5}\right)=\frac{1}{61}73^{s/2}*2\text{^}\left(s-\frac{8}{27 \sqrt{3} \log (3)}\right)\approx0.528786$$ and an upper bound of $$\frac{1}{122} 4^4 \left(\frac{\sqrt{876}}{81}\right)^s=\frac{1}{61}73^{s/2}*2^s\approx0.589052\text.$$
In short, if the horizontal base for the curve is $1\mathrm{m}$, and this paper is correct, then the $s$-dimensional Hausdorff measure (which is the appropriate way to measure the size of something a self-similar fractal like this) is definitely less than $0.6\mathrm{m}^{s}$, and probably more than $0.5\mathrm{m}^{s}$.