I found this basic integral in a stochastic calculus textbook:
$$ \int_0^t B_s \, dB_s = \frac{1}{2}B_t^2 - \frac{1}{2}t $$
where $B_t$ is Brownian motion. The RHS seems rather easy to understand, since we can get a PDF:
$$ \mathbb{P}\Big[\tfrac{1}{2}B_t^2 - \tfrac{1}{2}t \in [a,b] \Big] = \mathbb{P}\Big[B_t \in [\sqrt{2a+t},\sqrt{2b+t}] \Big] $$
This is not unlike a Beseel process...
Perhaps we could discretize the Brownian motion as a random walk (one of many possibilities): $$ B_s = \sum_{i=0}^s X_i $$
Can we do a similar "integration" except here it is just sum, and get a nice answer in the RHS?
$$ \sum_{n=0}^N\left(\sum_{i=0}^n X_i\right) X_i $$
where $X_i$ is a Bernoulli random variable with $\mathbb{P}[X_i = 0] = \mathbb{P}[X_i = 1] = \frac{1}{2}$.
Can I show this sum has a scaling limit to the Itô integral above?
You write this a bit incorrect (I don't mean the obvious misprint $X_n\to X_i$). The sum you wrote is a "backward" sum which involves the value $X_0 + \dots + X_n$ of the sum at "the right point" of "interval" $[n-1,n]$; so it corresponds to the so-called backward integral, not to the Itô integral. In order to recover the latter, you need to use the value "at the left" point, that is $X_0+\dots+X_{n-1}$.
That said, $$ \sum_{n=1}^N \left(\sum_{i=0}^{n-1} X_i\right) X_n = \frac12 \left(\sum_{n=0}^{N} X_n\right)^2 - \frac12 \sum_{n=0}^{N} X_n^2 = \frac{1}2\left(\sum_{n=0}^{N} X_n\right)^2 - \frac{N}2. $$ In particular, using the usual scaling by $\sqrt{N}$, $$ \sum_{n=1}^N \left(\sum_{i=0}^{n-1} \frac{X_i}{\sqrt{N}}\right) \frac{X_n}{\sqrt{N}} = \frac{1}2\left(\sum_{n=0}^{N} \frac{X_n}{\sqrt{N}}\right)^2 - \frac{1}2, $$ which does converge to the desired equality $$ \int_0^1 B_t\, dB_t = \frac12 B_1^2 -\frac{1}2. $$