There are many open questions related to the sharpness of Marstrand's slicing thorem, and some of them are actively studied by the community. Looking at the article linked by D.R., and on articles citing that article, is a good place to start. See also Chapter 6 in Mattila's 2015 book "Fourier Analysis and Hausdorff Dimension".
Regarding your specific questions, it is important to get the statement of Marstrand's slicing theorem right: fixing direction $\theta$ and taking $\ell_x=span(\theta)+x$ for $x\in\theta^\perp$, the estimate $$dim_H(S\cap\ell_x)\le dim_H(S)-1\quad\quad(*)$$ holds for Lebesgue almost every $x\in\theta^\perp$. The theorem is false if we replace "Lebesgue almost every" with "every".
In particular, if you consider the affine Grassmannian $A(2,1)$, which is a 2-dimensional manifold, then the estimate (*) is true for a.e. line $\ell\in A(2,1)$ (where a.e. refers to the natural Lebesgue measure on $A(2,1)$.
Regarding Question 1. If you ask for a lower bound on $dim_H(S\cap\ell)$ that would hold for a.e. $\ell$ in $A(2,1)$, then only the trivial bound $dim_H(S\cap\ell)\ge 0$ holds. One silly example: if $S\cap\ell=\varnothing$, which is true for the vast majority of lines if $S$ is compact, then $dim_H(S\cap\ell)= 0$. Assuming $S\cap\ell\neq\varnothing$ does not help much: if $S$ is a union of a line and a square, and $\ell$ intersects only the line, then we still get $dim_H(S\cap\ell)=0 < 1=dim_H(S)-1$. You can easily arrange for this to happen for a positive proportion of lines in a positive proportion of directions - so for a positive measure subset of $A(2,1)$.
On the other hand, as pointed out by D.R., Marstrand also proved that for a.e. direction $\theta$ there is a positive measure set of $x$ such that $dim_H(S\cap\ell_x)=dim_H(S)-1$, where again $\ell_x=span(\theta)+x$. So you know that the converse to (*) holds for a positive proportion of lines in $A(2,1)$ - but not for a.e. line. This leads us to question 2.
Regarding Question 2. A quite satisfactory answer can be found in the aforementioned book "Fourier Analysis and Hausdorff Dimension", Theorem 6.9., which in your setting says that if $0<H^s(S)<\infty$ for $1<s\le 2$, then there exists an exceptional set of directions $E\subset \mathbb{S}^1$ with $dim_H(E)\le 2-s$ such that for $H^s$-a.e. $x\in S$ and every $\theta\notin E$ we have
$$ dim_H(E\cap \ell_{x,\theta})= dim_H(E) - 1,$$
where $\ell_{x,\theta}=x+span(\theta)$.
It is worth stressing, that this result tells you that the converse to (*) holds for "a.e." line, but the meaning of "a.e." is different than we discussed in the answer to Question 1 - it does not refer to the Lebesgue measure on $A(2,1)$, but rather to "a.e." line passing through $E$, and "a.e." refers to the Hausdorff measure on $E$.
Best Answer
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.