a family F as above **of minimal size** satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details.
$$cov(\lambda,\theta,\kappa,\sigma):=\min\left\{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)\right\}$$
The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: http://papers.assafrinot.com/?num=5 .
Let $\lambda$ denote a singular cardinal. It always the case that $\lambda^+\le cov(\lambda,\lambda,cf(\lambda),2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the **least** (singular) cardinal for which any of the three is greater than $\lambda^+$, all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.