$ACC^{0}$ vs Poly-size circuits of bounded degree
$ACC^0$ and $SIZE(n^k)$ are probably incomparable for all $k$.

$ACC^0$ is conjectured not to contain $Maj(x_1,..,x_n)$, which has linear size circuits (think about the naive Turing machine for majority and transform it into a circuit, things are much easier when you don't have a depth restriction). This shows that it probably holds that $SIZE(n)\not\subseteq ACC^0$.

On the other hand, $ACC^0$ does allow you to use polynomially many gates (with no degree restriction). We will show that under a certain assumption, for every $k\in\mathbb{N}$ it holds that $ACC^0\not\subseteq SIZE(n^k)$. Consider the problem of deciding whether a graph contains a $c$-size clique (where $c$ is a constant independent of the input).

This problem is obviously in $AC^0$ (for all $c\in\mathbb{N}$), since you can, in the first level, link all subsets of $c$ vertices into AND gates to check whether they form a clique, and in the second level link all the results from the previous level into an OR gate.

If this can be done using $SIZE(n^k)$ circuits, then $\mathsf{NP\subseteq P/Poly}$, since we can construct polynomial size circuits for CLIQUE. The circuit for $n$ vertices graph needs to be able to check the existence of an $1,2,...,n$ clique. Since finding $c$-cliques is in $AC^0$ for all $c$, and we assumed $AC^0\subseteq SIZE(n^k)$, we have an $n^k$ size circuit for each possible clique size, combining those will result in an $n^{k+1}$ size circuit for clique on $n$ vertices graphs, which puts $\mathsf{NP}$ in $\mathsf{P/Poly}$.