Whether to use the factor $1/V$ or not gives different quantities. The difference is whether we talk about the free energy per volume (an intensive quantity, also called the free energy density) or the free energy of the whole system (an extensive quantity). The physical content of both is the same, so it is just a convention which we use.
In the following I work in units where $k = 1$, so $\beta = 1/T$.
The proper free energy contains a factor $T$: $\Phi = -T \log(Z)$. The missing $T$ can be explained by the fact, that the partition function is $Z = \mathop{\rm Tr}e^{-\beta H}$ and if we now have some Hamiltonian parametrized by a coupling strength $J$, so that $H = J \tilde H$, where $\tilde H$ is independent of $J$, we can see that the partition function only depends on $\beta J = J/T$. So we can simplify the problem setting $T = 1$ and tune our system with the parameter $J$ alone (of course this comes at a price: some observables become more cumbersome to compute, for example heat capacities).
Finally, I will show a way to derive $\Phi = -T\log(Z)$, using the canonical density operator $\rho = \frac 1 Z \mathop{\rm Tr}e^{-\beta H}$ and the definition of the entropy $S = -\mathop{\rm Tr}\big(\rho \log(\rho)\big)$:
\begin{align*}
\Phi &= E - TS = \mathop{\rm Tr}\left(\rho H\right) + T \mathop{\rm Tr}\big(\rho \log(\rho)\big) \\
&= \frac 1 Z \mathop{\rm Tr}(e^{-\beta H}H) + \frac T Z \mathop{\rm Tr}\big(e^{-\beta H} \log(e^{-\beta H}) - e^{-\beta H} \log(Z)\big) \\
&= \frac 1 Z \mathop{\rm Tr}(e^{-\beta H}H) - \frac T Z \beta \mathop{\rm Tr}(e^{-\beta H} H) - \frac T Z \mathop{\rm Tr} e^{-\beta H} \log(Z) \\
&= - T \log(Z).
\end{align*}
