So I have a filter $$H(z) = 0.5 + 0.5z^3 = (1/2, 0, 0, 1/2)$$ and need to find the output of it on a cyclical signal
$$x = (..., 3, -1, 2, 1, 5, 2, 3,-1, 2, 1, 5, 2, 3,...) $$
Would the output be simply the convolution of the coefficients
$$y = h*x $$
where x is flipped and moved across the filter?
Yes. You need to think of the convolution filter coefficients as a finite sequence, $h(k)=1/2$ if $t \in \{0,3\}$ and $h(k)=0$ elsewhere.