If $G$ is an adjoint algebraic group then it is always simple as an abstract group, (EDIT: This is because any proper normal subgroup of a simple algebraic group must be finite and lie in the centre). In general assume $G$ is connected reductive algebraic group and let $\pi : G \to G_{ad}$ be an adjoint quotient of $G$. Then $G$ is simple as an abstract group if the kernel of $\pi$ is trivial because $\pi$ is then an isomorphism of abstract groups. If $\pi$ does not have a trivial kernel then it is clear that $G$ is not simple as an abstract group.
For an example: in characteristic 2 the symplectic group $Sp_{2n}$ is simple as an abstract group as it is isomorphic (as an abstract group) to the corresponding adjoint group. However these two groups are not isomorphic as algebraic group (only isogenous).
EDIT: I suppose this didn't really explicitly answer your question. Explicitly we have a simple simply connected group is simple as an abstract group if and only if it is on the following list:
- $G_2$, $F_4$ or $E_8$ any characteristic.
- $B_n$, $C_n$, $D_n$ (n>2) or $E_7$ in characteristic 2.
- $E_6$ in characteristic 3.
- $A_n$ if $n$ is a power of the characteristic.