The primitive classes are the highest weight vectors.

Hard Lefschetz says that the operator $L$ (which algebraic geometers know as intersecting with a hyperplane) is the "lowering operator" $\rho(F)$ in a representation $\rho
\colon \mathfrak{sl}_2(\mathbb{C})\to End (H^\ast(X;\mathbb{C}))$. The raising operator $\rho(E)$ is $\Lambda$, the restriction to the harmonic forms of the the formal adjoint of $\omega \wedge \cdot$ acting on forms. The weight operator $\rho(H)$ has $H^{n-k}(X;\mathbb{C})$ as an eigenspace (= weight space), with eigenvalue (=weight) $k$.

The usual picture of an irreducible representation of $\mathfrak{sl}_2(\mathbb{C})$ is of a string of beads (weight spaces) with $\rho(F)$ moving you down the string and decreasing the weight by 2, and $\rho(E)$ going in the opposite direction. The highest weight is an integer $k$, the lowest weight $-k$.

From this picture, it's clear that the space of highest weight vectors in a (reducible) representation is $\ker \rho(E)$. It's also clear that, of the vectors of weight $k$, those which are highest weights are the ones in $\ker \rho(F)^{k+1}$. So the highest weight vectors in $H^{n-k}(X; \mathbb{C})$ are those in $\ker L^{k+1}$.

Of course, all this ignores the rather subtle question of how to explain in an invariant way what this $\mathfrak{sl}_2(\mathbb{C})$, or its corresponding Lie group, really is.

**Added**, slipping Mariano an envelope. But here's what that group is. Algebraic geometers, brace yourselves. Fix $x\in X$, and let $O_x = O(T_x X\otimes \mathbb{C})\cong O(4n,\mathbb{C})$. Then $O_x$ acts projectively on $\Lambda^\bullet (T_x X\otimes \mathbb{C})$ via the spinor representation (which lives inside the Clifford action). The holonomy group $Hol_x\cong U(n)$ also acts on complex forms at $x$, and the "Lefschetz group" $\mathcal{L}$ is the centralizer of $Hol_x$ in $O_x$. One proves that $\mathcal{L}\cong GL(\mathbb{C}\oplus \mathbb{C})$. Not only is this the right group, but its Lie algebra comes with a standard basis, coming from the splitting $T_x X \otimes\mathbb{C} = T^{1,0} \oplus T^{0,1}$. Now, $\mathcal{L}$ acts on complex forms on $X$, by parallel transporting them from $y$ to $x$, acting, and transporting back to $y$. Check next that the action commutes with $d$ and $*$, hence with the Laplacian, and so descends to harmonic forms = cohomology. Finally, check that the action of $\mathcal{L}$ exponentiates the standard action of $\mathfrak{gl}_2$ where the centre acts by scaling. (This explanation is Graeme Segal's, via Ivan Smith.)