Let $X = Hom(T, G_m)$ and $Y = Hom(G_m, T)$ be the character and cocharacter lattices of $T$, respectively. Let $k$ be the (algebraically closed) ground field. Note that $T = Spec(k[X])$ (a canonical identification).

Let $Z$ be the center of $G$. Then $Z = Spec(k[X / R])$ where $R$ is the root lattice in $X$. On the other hand, $T^W = Spec(k[X_W])$, where $X_W$ denotes the $W$-coinvariants, i.e.,
$$X_W = X / \langle x - wx : x \in X, w \in W \rangle.$$
Let $D = \langle x - wx : x \in X, w \in W \rangle \subset X$. Note that $D$ is certainly contained in $R$, since $x - wx$ will always be a sum of roots. Tracking things through, the inclusion $D \hookrightarrow R$ corresponds to a surjection $X / D \twoheadrightarrow X/R$ and via $Spec(K[\bullet])$ to the inclusion $Z \hookrightarrow T^W$.

The short exact sequence
$$1 \rightarrow R/D \rightarrow X/D \rightarrow X/R \rightarrow 1$$
corresponds to a short exact sequence of groups of multiplicative type,
$$1 \rightarrow Z \rightarrow T^W \rightarrow Spec(k[R/D]) \rightarrow 1.$$
So you can see that the deviation between $Z$ and $T^W$ is precisely measured by $R/D$.

Note that $D$ contains $2 \alpha$ for every root $\alpha$ (note $x - wx = 2 \alpha$ if $x = \alpha$ and $w = s_\alpha$). Hence $D$ is a full-rank sublattice of $R$, and so $R/D$ is finite. It follows that the homomorphism $Z \rightarrow T^W$ restricts to an isomorphism on neutral components.

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Let's look at $D$ a bit more closely, to determine when $D = R$. Fix a system of simple roots $\Delta$. Then
$$D = \langle x - wx : x \in X, w \in W = \langle x - s_\alpha(x) : x \in X, \alpha \in \Delta \rangle.$$
But $x - s_\alpha(x) = \langle x, \alpha^\vee \rangle \alpha$. So while the root lattice is given by $R = {\mathbb Z} \Delta$, the sublattice $D$ is generated by $\langle x, \alpha^\vee \rangle \alpha$ for all $\alpha \in \Delta$.

Let $G'$ be the derived subgroup, and $T' = T \cap G'$, with cocharacter lattice $Y'$ and character lattice $X'$. Then $Y' = Y \cap (R^\vee \otimes {\mathbb Q})$, where $R^\vee$ is the coroot lattice, and
$$X' = X / \{ x \in X : \forall \alpha \in \Delta, \langle x, \alpha^\vee \rangle = 0 \}.$$

Thus $D$ only depends on $G'$.

$$D = \langle \langle x, \alpha^\vee \rangle \alpha : x \in X', \alpha \in \Delta \rangle.$$
Now $D = R$ if and only if for all $\alpha^\vee$, the exists $x \in X'$ such that $\langle x, \alpha^\vee \rangle = 1$. Now what to do...

Decompose the root system into irreducible subroot systems, yielding $\Delta = \Delta_1 \sqcup \cdots \Delta_m$. Suppose that $\alpha \in \Delta_j$ and $\Delta_j$ does not have type $B$. Avoid $A_1 = B_1$ too! Then, there exists a root $\beta$ such that $\langle \beta, \alpha^\vee \rangle$ is odd. It follows that there exists $x \in X'$ such that $\langle x, \alpha^\vee \rangle = 1$.

On the other hand, if $\Delta_j$ does have type $B$ (or type $A_1 = B_1$), then $R = D$ implies the existence of $x \in X'$ such that $\langle x, \alpha^\vee \rangle = 1$ for the unique long simple coroot $\alpha^\vee$.

So, what does this all mean? Consider the simply-connected cover of $G'$, which fits into a short exact sequence,
$$1 \rightarrow \pi_1(G') \rightarrow G_{sc} \rightarrow G' \rightarrow 1.$$

The simply-connected group $G_{sc}$ is a direct product of simple factors; I think that $R = D$ is equivalent to the statement that for every type $B$ factor $Spin_{2n+1}$ in $G_{sc}$, the induced map $Spin_{2n+1} \rightarrow G'$ is injective (i.e., it doesn't factor through $SO_{2n+1}$). I haven't checked the details... but it seems consistent with Kasper Andersen's discussion for compact Lie groups.