I'm not saying the distribution of $L$ is inappropriate, but I think it will be more easy to work with another one.

Let me give an answer that works with a general class of distributions. I also only assume that the tessellation is only made from convex polytopes.

First remark that if I denote by $T$ the union of all faces of cells of the tessellation (edges of $[0,1]^n$ included), you are interested in the variable $$F=card (T\cap L)-1$$ (because a.s. each time the line touches a cell, it only touches its border twice, and the exit point is actually the entrance point for a new cell, except at the end-points).

Let $\mathcal{L}$ be the class of all lines of $\mathbb{R}^n$, and $\mu$ some measure on $\mathcal{L}$ (endowed with a topology coming from a parametrization of $\mathcal{L}$), and for a set $C$ of $\mathbb{R}^n$, denote by $[C]$ the class of all lines intersecting $C$. Instead of working with a single line $L$, consider an independant Poisson point process of lines $\Pi$ on $\mathcal{L}$ with intensity measure $\mu$. I emphasize that a.s. only a finite number of lines of $\Pi$ will hit $[0,1]^n$, so I can label them independantly $N_1, N_2, ...$ so that the $L_i$ are iid. Call $N$ then number of lines touching $[0,1]^n$. Given any fixed tessellation $T$, we have $$\mathbb{E}\sum_i card(T\cap L_i)=\mathbb{E}\sum_n \mathbb{P}(N=n) \sum_{i=1}^n card(T\cap L_i)=\sum_n \mathbb{P}(N=n) n\mathbb{E} card(T\cap L_1)$$

$$=(\mathbb{E}(F)+1)\mathbb{E}(N).$$

So we will compute $\mathbb{E}(N)$ and $\mathbb{E}\sum_i card(T\cap L_i)$. We have $\mathbb{E}(N)=\mu([[0,1]^n])$ because $\Pi$ is a Poisson point process.

Let $T$ be a deterministic tessellation, written as $T=\cup f$, where the $f$ are the facets of the tessellations. Remarking that any given line can hit a facet no more than one time, we have $$\mathbb{E}\sum_i card(T\cap L_i)=\mathbb{E}\sum_i\sum_f card(f\cap L_i)=\sum_f \mathbb{E} card(i:f\cap L_i\neq \emptyset)=\sum_f \mu([f])$$ because $(L_1,L_2,...)$ is a Poisson point process with intensity $\mu$.

At this point you need to make assumptions on the distribution of the lines, so I make the assumption that $\mu$ is translation invariant. In this case it is a standard fact from integral geometry that $\mu([f])=c |f|$ where $c$ is a constant depending solely on $\mu$ and $|f|$ is the $(n-1)$-dimensional measure of $f$. Thus you have $$\mathbb{E}F=\mathbb{E}\frac{\sum_f c|f| }{\mu([[0,1]^n])}-1=c\frac{\mathbb{E}|T|}{\mu([[0,1]^n])}-1.$$

With normalisation conditions you can probably compute $c/\mu([[0,1]^n])$. Thus the above results holds for any random tessellation with facets as convex polytopes and a random line segment that is the restriction of a stationary measure $\mu$ on $\mathcal{L}$ to $[[0,1]^n$. For example the uniform distribution works, but you can also choose a different directional distribution, for instance "only horizontal lines and vertical lines", etc...I don't know if the iid intersection points enters this framework.