No. Consider the 2-category $C = [2 \mid 1, 1]$, which has 3 objects $0,1,2$; 4 generating 1-morphisms $f,f' : 0 \rightrightarrows 1$ and $g,g' : 1 \rightrightarrows 2$; and 2 generating 2-cells $\alpha : f \Rightarrow f'$ and $\beta : g \Rightarrow g'$. $C$ arises from a computad.
Let $i=0$. Then $C^{(0)}$ is the disjoint union of the following 6 categories:
3 copies of the terminal category $[0]$ given by $id_0, id_1, id_2$;
2 copies of the arrow category $[1]$ given by $f \xrightarrow \alpha f'$ and $g \xrightarrow \beta g'$;
1 copy of the square category $[1] \times [1]$, given by $gf \overset{g\alpha, \beta f}{\rightrightarrows} gf', g'f \overset{\beta f', g' \alpha}{\rightrightarrows} g'f'$ (the diagonal is $\beta \alpha$).
1 copy of the square category $[1] \times [1]$, given by $gf \overset{g\alpha}{\rightarrow} gf' \overset{\beta f'}{\rightarrow} g'f'$, $gf \overset{\beta f}{\rightarrow} g'f \overset{g' \alpha}{\rightarrow} g'f'$ (the diagonal is $\beta \alpha$).
This last factor has 4 atomic 1-morphisms on the outside of the square; if $C^{(0)}$ were a computad, then we would have $\beta f' \circ g \alpha \neq g'\alpha \circ \beta f$, but in fact these composites are both equal to $\beta \alpha$. So $C^{(0)}$ is not a computad.
