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, $M$ and $M^{\perp}$ generate $L_{K3}$, so this gives the whole of $H^2(K3,\mathbb Z)$.
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 the lattice $M$ described above and the image of the Picard group of $A$.