For brevity, call a "wall" built with $n$ cubes an $n$-wall. Call two $n$-walls different if, viewed from left to right, they look different. We give two ways of counting the $n$-walls, the second much simpler than the first. But the first way has the advantage that it arises from experimentation with small numbers. That always should be done, in order to try to get insight about a new combinatorial problem.
It is clear that the number of $1$-walls is $1$.
The $2$-walls are $11$ and $2$. There are $2$ of them.
The $3$-walls are $111$, $21$, $12$, and $3$. There are $4$ of them.
We list the $4$-walls, carefully. There are $8$ of them, $1111$, $211$, $121$, $31$, $112$, $22$, $13$, and $4$.
It is not unreasonable to conjecture that the number of $n$-walls is $2^{n-1}$, though admittedly the evidence is scanty.
Now let's prove the conjecture, technically by induction. Suppose that we know for some specific $n$, that the number of $n$-walls is $2^{n-1}$. How many $(n+1)$-walls are there?
There are two types of $(n+1)$-wall. Type $A$ has a $1$ at the right end. Type $B$ has a number $>1$ at the right end.
It is obvious that the number of type $A$ $(n+1)$-walls is $2^{n-1}$. The type $A$ $(n+1)$-walls are obtained by taking any $n$-wall, and appending a $1$ at the right end. Given any type $A$ $(n+1)$-wall, we can uniquely recover the $n$-wall that it "came from" by removing the terminal $1$.
There are also $2^{n-1}$ type $B$ $(n+1)$-walls. They are obtained by adding a cube on top of the rightmost pile of an $n$-wall. Given a type $B$ $(n+1)$-wall, we can uniquely recover the $n$-wall that it "came from" by removing the top cube from the rightmost pile of the type $B$ $(n+1)$-wall.
It follows that the number of $(n+1)$-walls is $2^{n-1}+2^{n-1}$, that is, $2^n$, twice as many as the number of $n$-walls.
Since we know that the number of $1$-walls is $2^{1-1}$, this completes the induction process. More informally, we know that there is $1$ $1$-wall. Therefore, without listing, we know that there must be $2$ $2$-walls. Therefore, without listing, we know there must be $4$ $3$-walls. But therefore there must be $8$ $4$-walls. And therefore there must be $16$ $5$-walls. And so on forever.
Another way: Imagine that the child's older sister is a professional wall designer. She lays out the $n$ cubes in a long row, with a space of a few centimeters between consecutive cubes. There are $n$ cubes, and therefore $n-1$ spaces.
The designer then chooses a subset of the set of $n-1$ spaces, and places a ribbon at each chosen space. (She might choose the empty subset of the $n-1$ spaces, or just $1$ of the spaces, or $2$, and so on.)
Now her kid brother builds the wall. The cubes from the left end of the row to the first ribbon are piled on each other to make the leftmost part of the wall (if $0$ spaces were chosen, all the cubes go on top of each other). The cubes from the first ribbon to the second are piled on top of each other, immediately to the right of the first pile. And so on.
It is clear that there are exactly as many $n$-walls as there are ways to choose a subset of the $n-1$ gaps. But any set of $n-1$ elements has $2^{n-1}$ subsets, so there are $2^{n-1}$ $n$-walls.
Comment: Let's add a rule to the wall building: The wall must be exactly $m$ cubes long. Then in the second argument, the designer must choose exactly $m-1$ of the $n-1$ gaps to put a ribbon into. It follows that the number of different walls of length exactly $m$ is $\binom{n-1}{m-1}$.