Let $X=(X_n)_{n\in\mathbb{N}}$ be a sequence of independent identically distributed random variables satisfying $\mathbb{E}X_1^2=1$. Let $\mathbb{F}=(\mathcal{F}_n)_{n\in\mathbb{N}}$ the filtration generated by X. (Or $\mathcal{F}_n=\sigma(X_1,X_2,\dots,X_n)$.)
Why is $\mathbb{E}[X_n^2|\mathcal{F}_{n-1}]=1$?
If it follows from independence of $X_n$ and $\mathcal{F}_{n-1}$, what is the intuition behind it? Or does it follow from the fact that $X$ is iid? I would like a detailed answer so I can really understand what is happening here and be able to recognise similar situations in the future.
It follows from independence.
If $X$ and $Y$ are random variables then we can go searching for $Z=\mathbb E[X\mid Y]$ (also denoted as $\mathbb E[X\mid \sigma(Y)]$) which is a random variable that is characterized by:
The first bullet can be restated as: $Z=f(Y)$ for some Borel-measurable function $f:\mathbb R\to\mathbb R$.
The second bullet implies that $\mathbb EXg(Y)=\mathbb EZg(Y)$ for every Borel-measurable function $g$.
In the special case where $X$ and $Y$ are independent we can take for $Z=\mathbb E[X\mid Y]$ the constant function that is prescribed by: $\omega\mapsto\mathbb EX$.
As a constant function of course it is measurable wrt $\sigma(Y)$ (so the first bullet is satisfied).
Concerning the second bullet note that in this case $X$ and $\mathbf1_A$ are independent so that: $$\mathbb EX\mathbf1_A=\mathbb EX\cdot\mathbb E\mathbf1_A=\mathbb E[[\mathbb EX]\mathbf1_A]$$
So in your case we get:$$\mathbb{E}[X_n^2|\mathcal{F}_{n-1}]=\mathbb{E}[X_n^2]=1$$