5 Added "Kummer lattice" and a remark about the cone of effective curves.; added 7 characters in body; edited body

The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is $2E_8+3U$, with $E_8$ negative definite and $U$ the hyperbolic lattice for the bilinear form $xy$. It is unimodular and has signature $(3,19)$.

The 16 (-2)-curves $E_i$ form a sublattice $16A_1$ of determinant $2^{16}$. It is not primitive in $L_{K3}$. The primitive lattice $M$ K$containing it is computed as follows. Consider a linear combination$F=\frac12\sum a_i E_i$with$a_i=0,1$. Recall that$E_i$are labeled by the 2-torsion points of the torus$A$, i.e. the elements of the group$A[2]$. Then$F$is in$M$K$ $\iff$ the function $a:A[2]\to \mathbb F_2$, $i\mapsto a_i$, is affine-linear. You will find the proof of this statement in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5. (The element $\frac12\sum E_i$ in your example corresponds to the constant function 1, which is affine linear). Thus, $M$ K$has index$2^5$in$16A_1$and its determinant is$2^{16}/(2^5)^2=2^6$.$K$is called the Kummer lattice. By the above, it is a concrete negative-definite lattice of rank 16 with determinant$2^6$. Nikulin proved that a K3 surface is a Kummer surface iff$Pic(X)$contains$K$. The orthogonal complement$M^{\perp}$K^{\perp}$ of $M$ K$in$L_{K3}$is$H^2(A,\mathbb Z)$but with the intersection form multiplied by 2. As a lattice, it is isomorphic to$3U(2)$. It has determinant$2^6$, the same as$M$. K$. The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is recovered from the primitive orthogonal summands $M$ K$and$M^{\perp}$.K^{\perp}$.

However, your question has "Picard lattice" in the title. The Picard group of $X$ is strictly smaller than $H^2(X,\mathbb Z)$. To begin with, it has signature $(1,r)$, (1,r-1)$, not$(3,19)$. For the a Kummer surface, it contains the Kummer lattice$M$K$ described above, and its intersection with $M^{\perp}$ K^{\perp}$is the image of the Picard group of$A$. For a Kummer surface one has$r=17,18,19$or 20. For the Mori-Kleiman cone of effective curves, which you would need for Gromov-Witten theory, the description you put in a box is already the best possible. 4 added 44 characters in body The lattice$L_{K3}=H^2(K3,\mathbb Z)$is$2E_8+3U$, with$E_8$negative definite and$U$the hyperbolic lattice for the bilinear form$xy$. It is unimodular and has signature$(3,19)$. The 16 (-2)-curves$E_i$form a sublattice$16A_1$of determinant$2^{16}$. It is not primitive in$L_{K3}$. The primitive lattice$M$containing it is computed as follows. Consider a linear combination$F=\frac12\sum a_i E_i$with$a_i=0,1$. Recall that$E_i$are labeled by the 2-torsion points of the torus$A$, i.e. the elements of the group$A[2]$. Then$F$is in$M\iff$the function$a:A[2]\to \mathbb F_2$,$i\mapsto a_i$, is affine-linear. You will find the proof of this statement in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5. (The element$\frac12\sum E_i$in your example corresponds to the constant function 1, which is affine linear). Thus,$M$has index$2^5$in$16A_1$and its determinant is$2^{16}/(2^5)^2=2^6$. The orthogonal complement$M^{\perp}$of$M$in$L_{K3}$is$H^2(A,\mathbb Z)$but with the intersection form multiplied by 2. As a lattice, it is isomorphic to$3U(2)$. It has determinant$2^6$, the same as$M$. Together, The lattice$L_{K3}=H^2(K3,\mathbb Z)$is recovered from the primitive orthogonal summands$M$and$M^{\perp}$generate$L_{K3}$, so this gives the whole of$H^2(K3,\mathbb Z)$.M^{\perp}$.

However, your question has "Picard lattice" in the title. The Picard group of $X$ is strictly smaller than $H^2(X,\mathbb Z)$. To begin with, it has signature $(1,r)$, not $(3,19)$. For the Kummer surface, it is generated by contains the lattice $M$ described above, and its intersection with $M^{\perp}$ is the image of the Picard group of $A$.

The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is $2E_8+3U$, with $E_8$ negative definite and $U$ the hyperbolic lattice for the bilinear form $xy$. It is unimodular . and has signature $(3,19)$.
The 16 (-2)-curves $E_i$ form a sublattice $16A_1$ of determinant $2^{16}$. It is not primitive in $L_{K3}$. The primitive lattice $M$ containing it is computed as follows. Consider a linear combination $F=\frac12\sum a_i E_i$ with $a_i=0,1$. Recall that $E_i$ are labeled by the 2-torsion points of the torus $A$, i.e. the elements of the group $A[2]$.
Then $F$ is in $M$ iff $\iff$ the function $a:A[2]\to \mathbb F_2$, $i\mapsto a_i$, is affine-linear. You will find the proof of this statement in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5. (The element $\frac12\sum E_i$ in your example corresponds to the constant function 1, which is affine linear). Thus, $M$ has index $2^5$ in $16A_1$ and its determinant is $2^{16}/(2^5)^2=2^6$.
The orthogonal complement $M^{\perp}$ of $M$ in $L_{K3}$ is $H^2(A,\mathbb Z)$ but with the intersection form multiplied by 2. As a lattice, it is isomorphic to $3U(2)$. It has determinant $2^6$, the same as $M$. Together, $M$ and $M^{\perp}$ generate $L_{K3}$, so this gives the whole of $H^2(K3,\mathbb Z)$.(All of this is explained in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5.)
However, your question has "Picard lattice" in the title. The Picard group of $X$ is strictly smaller than $H^2(X,\mathbb Z)$. To begin with, it has signature $(1,r)$, not $(3,19)$. For the Kummer surface, it will contain is generated by the lattice $M$ described above and the image of the Picard group of $A$.