Let $\ell\ge0$ be the characteristic of the (algebraically closed) ground field. Let $G$ be semisimple with Lie algebra $\mathfrak g$.

First, the maximal dimension of an abelian subalgebra of $\mathfrak g$ can increase for small $\ell$. Let, e.g., $\mathfrak g=sl(2)$ and $\ell=2$. Then the Borel subalgebra is abelian, hence the maximal dimension is $2$ instead of $1$. The same holds for $\mathfrak g=pgl(2)$ and $\ell=2$ where $\langle e,f\rangle$ is abelian.

Secondly, this is a phenomenon of very small characteristics. The following statement is not the most general one but covers most cases:

Assume that $G$ is of adjoint type and $\ell>3$. Then the dimension of a maximal abelian subalgebra is the same as in characteristic $0$.

Proof: Key is the following deformation argument: Being an abelian subalgebra is a closed condition in the Grassmannian of $\mathfrak g$. In other words, there is a closed subscheme $A_d\subseteq Gr_d(\mathfrak g)$ classifying abelian subalgebras of dimension $d$. The group $G$ acts on $A_d$ by conjugation. Assume that $A_d\ne\emptyset$. Since $A_d$ is projective, any Borel subgroup $B$ of $G$ has a fixed point. This shows, that if $\mathfrak g$ contains an abelian subalgebra $\mathfrak a$ of dimension $d$ then it will also contain one which is normalized by $B$. Assume from now on that this is the case.

Then, in particular, $\mathfrak a$ will be normalized by a maximal torus $T\subseteq B$. Let $\mathfrak g=\mathfrak t\oplus\bigoplus_\alpha\mathfrak g_\alpha$ be the root space decomposition. Then any $T$-stable subspace, so also $\mathfrak a$, has the form
$$
\mathfrak a=\mathfrak a_0\oplus\bigoplus_{\alpha\in S}\mathfrak g_\alpha
$$
where $S$ is a set of roots.

Since $\mathfrak a$ is ablian, its subset of semisimple elements $\mathfrak a_0$ inside $\mathfrak b$ is also normalized by $B$. Hence $\mathfrak a_0$ consists of fixed points of $B$ and therefore of $G$. So $\mathfrak a_0$ sits in the schematic center of $\mathfrak g$. Because $G$ is of adjoint type we infer $\mathfrak a_0=0$.

Thus everything depends on $S$. I argue that the condition on $S$ making $\mathfrak a=\bigoplus_{\alpha\in S}\mathfrak g_\alpha$ abelian is independent of $\ell$, proving our assertion.

For this let $e_\alpha\in\mathfrak g_\alpha$ be a Chevalley generator. Then $[e_\alpha,e_\beta]$ must be zero for all $\alpha,\beta\in S$. If $\alpha+\beta=0$ then $h_\alpha=[e_\alpha,e_\beta]\ne0$ since $\ell\ne2$. So this case must not occur. If $\gamma=\alpha+\beta$ is a root then a famous formula of Chevalley asserts
$$
[e_\alpha,e_\beta]=\pm N_{\alpha\beta}e_\gamma\quad\text{with }N_{\alpha\beta}\in\{1,2,3\}.
$$
Because $\ell>3$, by assumption, we get $[e_\alpha,e_\beta]\ne0$. So this case must not occur either. In the remaining cases we have $[e_\alpha,e_\beta]=0$. The condition on $S$ is therefore: $\mathfrak a$ is an abelian subalgebra if and only if $\alpha+\beta$ is not zero and not a root for all $\alpha,\beta\in S$. This condition is clearly characteristic free.