Edit: I have elaborated on this approach to the Picard group in Section 2 of my preprint.

The question was answered in the comments above, but only for the case of torsion-free Picard group. However, for non-complete toric varieties, the Picard group may have torsion (see example below). Since this showed up in my research, let me describe a method of determining the Picard group of a normal toric variety, where the fan is not necessarily simplicial or complete. I will give a criterion for the Picard group to be torsion-free and describe how to determine its rank.

Let $X$ be a $n$-dimensional normal toric variety associated to a fan $\Sigma$ in a lattice $N\cong\mathbb{Z}^n$. Let $v_1, \dots, v_r\in N$ be generators of the rays (one-dimensional cones) of $\Sigma$ and write $F:=\mathbb{Z}^r$. I assume that $X$ has no torus factor, that means $v_1, \dots, v_r$ generate $N \otimes \mathbb{Q}$ as a vector space. We have a canonical map of lattices
$$
P \colon F \to N, \qquad e_i \mapsto v_i.
$$
Writing $E:=F^*$ and $M:=N^*$ for the dual lattices, standard toric geometry gives an exact sequence
$$\require{AMScd}
\begin{CD}
0 @>>> M @>{P^*}>> E @>>> \mathrm{Cl}(X) @>>> 0,
\end{CD}$$
where $\mathrm{Cl}(X)$ is the divisor class group (Weil divisors modulo principal divisors) of $X$. I view the Picard group $\mathrm{Pic}(X)$ as Cartier divisors modulo principal divisors, hence it is a subgroup of $\mathrm{Cl(X)}$. Note that a Weil Divisor on a toric variety is Cartier iff it is principal on all affine toric charts associated to the cones $\sigma \in \Sigma$.

So let's work with a single affine chart $U_\sigma$ for a cone $\sigma = \mathrm{poshull}(v_{i_1}, \dots, v_{i_{r_\sigma}}) \in \Sigma$. Define
$$
N(\sigma) := \mathrm{lin}_{N\otimes\mathbb{Q}}(\sigma) \cap N, \qquad F(\sigma) := \mathbb{Z}^{r_\sigma}.
$$
Note that $\dim(N(\sigma)) \leq r_\sigma$ and equality holds iff $\sigma$ is simplicial. Let $\alpha_\sigma\colon N(\sigma) \to N$ be the inclusion and $\beta_\sigma\colon F(\sigma) \to F, e_j \mapsto e_{i_j}$. Setting $M(\sigma) := N(\sigma)^*$ and $E(\sigma):= F(\sigma)^*$, we obtain a commutative diagram with exact rows
$$\require{AMScd}
\begin{CD}
0 @>>> M @>{P^*}>> E @>>> \mathrm{Cl}(X) @>>> 0 \\
@.@V{\alpha_\sigma^*}VV @V{\beta_\sigma^*}VV @V{\pi_\sigma}VV @. \\
0 @>>> M(\sigma) @>>> E(\sigma) @>>> \mathrm{Cl}(U_\sigma) @>>> 0.
\end{CD}$$
where $\pi_\sigma$ maps a divisor class $[D]$ to the restriction $[D|_{U_\sigma}]$. Being principal on $U_\sigma$ means being in the kernel of $\pi_\sigma$. For the Picard group, this means
$$
\mathrm{Pic}(X) = \bigcap_{\sigma \in \Sigma} \ker{\pi_\sigma}.
$$
We now draw the above diagram again, but for all cones $\sigma\in\Sigma$ at the same time. That is, we define
$$
\mathbf{F}:=\bigoplus_{\sigma\in\Sigma} F(\sigma),\quad \mathbf{N}:=\bigoplus_{\sigma\in\Sigma} N(\sigma),\quad \mathbf{M}:=\mathbf{N}^*,\quad\mathbf{E}:=\mathbf{F}^*.
$$
Furthermore, we obtain maps $\alpha\colon \mathbf{N} \to N, \beta\colon \mathbf{F} \to F$ and $\pi\colon \mathrm{Cl}(X) \to \bigoplus_{\sigma\in\Sigma} \mathrm{Cl}(U_\sigma)$ and we have $\mathrm{Pic}(X) = \ker{\pi}$. Note that $\beta$ is surjective, hence $\beta^*$ is injective. Furthermore, since $v_1, \dots, v_r$ generate $N \otimes \mathbb{Q}$ as a vector space, the cokernel of $\alpha$ is finite, so $\alpha^*$ is injective as well. We obtain a diagram with exact rows and columns
$$\require{AMScd}
\begin{CD}
@. 0 @. 0 @. \mathrm{Pic}(X) @. \\
@. @VVV @VVV @VVV @. \\
0 @>>> M @>{P^*}>> E @>>> \mathrm{Cl}(X) @>>> 0 \\
@.@V{\alpha^*}VV @V{\beta^*}VV @V{\pi}VV @. \\
0 @>>> \mathbf{M} @>>> \mathbf{E} @>>> \bigoplus_{\sigma\in\Sigma} \mathrm{Cl}(U_\sigma) @>>> 0. \\
@. @VVV @VVV @VVV @. \\
@.\mathrm{coker}(\alpha^*) @>>> \mathrm{coker}(\beta^*) @>>> \mathrm{coker}(\pi) @>>> 0 \\
\end{CD}$$
The snake lemma now identifies $\mathrm{Pic}(X)$ with a subgroup of $\mathrm{coker}(\alpha^*)$. This gives a criterion for torsion-freeness of the Picard group: If $\alpha$ is surjective, $\mathrm{coker}(\alpha^*)$ is torsion-free, hence so is the Picard group.

Note that as soon as $\Sigma$ contains a cone of maximal dimension (for instance if it is complete), this holds trivially, since then $N(\sigma)=N$. A counterexample is the fan $\Sigma$ in $\mathbb{Z}^2$ having as maximal cones the two rays generated by $(1,0)$ and $(1,2)$. The Picard group of the associated smooth toric surface has torsion, indeed it has $\mathrm{Pic}(X)=\mathrm{Cl}(X)\cong\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.

To determine the rank of the Picard group, look again at the exact sequence obtained by applying the snake lemma to the above diagram. Since the $\pi_\sigma$ are all surjective, the cokernel of $\pi$ is finite. Since the alternating sum of dimensions in an exact sequence of vector spaces vanishes, we obtain

$$
\begin{align*}
\mathrm{rank}(\mathrm{Pic}(X)) & = \mathrm{rank}(\mathrm{coker}(\alpha^*)) - \mathrm{rank}(\mathrm{coker}(\beta^*)) \\
& = r - n - \left(\sum_{\sigma\in\Sigma} r_\sigma - \dim(\sigma)\right)
\end{align*}
$$

Note that this is exactly the formula found in Ewald's *Combinatorial Complexity and Algebraic Geometry* mentioned in the comments above.

Combinatorial convexity and algebraic geometry,Springer GTM 168 (1996). $\endgroup$geometric quotientfrom the equivalence relation (as opposed to a categorical quotient in general). $\endgroup$4more comments