If $f$ and $g$ are both integrable and square-integrable, and $f$ is super-compact, and $f*g$ is continuous, then $f*g$ is super-compact. (This covers your specific case, at least if the compactly supported $f$ is continuous.) Note we don't even need $f$ and $g$ to be nonnegative.
We need to prove that the $t$-level sets are both closed and bounded. First, closed is easy: the $t$-level set is the inverse image of the closed interval $[t,\infty)$ under the map $f*g\colon \mathbb R^d\to\mathbb R$, and hence it is automatically closed (by the topological definition of continuity).
As for bounded: given $\varepsilon>0$, choose $R=R(\varepsilon)$ so large that $|f(x)|<\varepsilon$ for $|x|>R$ (by super-compactness) and that the integral of $|g(x)|^2$ over the complement of the ball of radius $R$ is less than $\varepsilon$ (by square-integrability). Then for $|x|>2R$, $ (f*g)(x) = \int_{\mathbb R^d} f(y)g(x-y) \,dy = \int_{|y|\le R} f(y)g(x-y) \,dy + \int_{|y|> R} f(y)g(x-y) \,dy. $ In the first integral, use Cauchy-Schwarz: \begin{align*} \bigg| \int_{|y|\le R} f(y)g(x-y) \,dy \bigg| &\le \bigg( \int_{|y|\le R} |f(y)|^2 \,dy \bigg)^{1/2} \bigg( \int_{|y|\le R} |g(x-y)|^2 \,dy \bigg)^{1/2} \\ &\le \bigg( \int_{|y|\le R} |f(y)|^2 \,dy \bigg)^{1/2} \bigg( \int_{|z|> R} |g(z)|^2 \,dy \bigg)^{1/2} < \|f\|_2 \varepsilon^{1/2}. \end{align*} (Here we've used the fact that $|x|>2R$ and $|y|\le R$ imply $|z|>R$ in the change of variables $z=x-y$.) For the second integral, $ \bigg| \int_{|y|> R} f(y)g(x-y) \,dy \bigg| < \varepsilon \int_{|y|> R} |g(x-y)| \,dy \le \|g\|_1 \varepsilon. $ Thus $|(f*g)(x)|$ is bounded by $\|f\|_2 \varepsilon^{1/2} + \|g\|_1 \varepsilon$ for all $|x|>2R(\varepsilon)$. In other words, the $t$-level set is bounded for all $t \ge \|f\|_2 \varepsilon^{1/2} + \|g\|_1 \varepsilon$, since $\varepsilon$ can be taken arbitrarily small, so can $t$.