I'm studying ordinal arithmetic and I want to understand numbers like $\omega + \omega^2 + \omega^3$ using queue order types. So the idea that's been explained to me is that $3+\omega$ is the "queue type" when you add three things to the start of a queue of length $\omega$, which is still just a single countably infinite queue, so $3+\omega = \omega$. Conversely, $\omega+3$ is a queue with three items after everything in the first countably infinite queue, which is different to $\omega$ (which has no largest element).
With this, I understand $\omega^k$ to be a queue $k$ layers deep, and when you "zoom in" this many times, you get a single object.
My intuition behind sums is then that, say, $\omega+\omega^2+\omega^3$ is a countable queue followed by countable queues containing countable queues, followed by countable queues of countable queues of countable queues. The first queues do not add any "waiting times" to anything in the last queue, so I would expect this to be equal to $\omega^3$.
As a concrete question, consider the ordinals $\alpha := \omega+\omega^2+\omega^3$, $\beta:=\omega^3+\omega+\omega^2$, $\gamma:=\omega^3+\omega^2+\omega$ and $\delta:= \omega^2+\omega+\omega^3$. Are any of these ordinals equal? For the ones which aren't (if any), order them in ascending order. Argue not with proofs, but with "queue size" intuition.
Your expectation that $\alpha=\omega^3$ is correct. The same idea works to see that $\delta=\omega^3$. Both $\beta$ and $\gamma$ tack shorter bits on after the initial $\omega^3$, just as $\omega+3$ tacks an extra $3$ things at the end of the $\omega$ queue, so you should expect that both are larger than $\omega^3$, and the question now is which of $\beta$ and $\gamma$ is larger.
The same heuristic that told you that $\alpha=\omega^3$ tells you that $\omega+\omega^2=\omega^2$, so $\beta$ just tacks an extra $\omega^2$ things after the initial $\omega^3$ queue; $\gamma$ then adds another $\omega$ things after that, so $\alpha=\delta<\beta<\gamma$.