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2 added 50 characters in body; added 58 characters in body

I am not sure whether this is exactly what you want. But still, I see the following relation between $Pic$ and $\pi_1$:

Let $k$ be an algebraically closed field and $X/k$ a proper variety. Then there is an isomorphism $H^1(X, \mathbb{Z}/n)=Hom(\pi_1(X), \mathbb{Z}/n)$. Furthermore, from the long exact cohomology sequence associated to $$0\to \mathbb{Z}/n\to \mathbb{G}_m\to \mathbb{G}_m\to 0$$ we obtain an isomorphism $$H^1(X, \mathbb{Z}/n)\cong H^1(X, \mathbb{G}_m)[n].$$ Here we used that $k$ is algebraically closed and that $X$ is proper over $k$. By a generalization of Hilbert 90 we have $H^1(X, \mathbb{G}_m)=Pic(X)$. (Cf. Milne's book, Chapter III, Proposition 4.9.) Hence, after all, we see that there is an isomorphism $$Hom(\pi_1(X), \mathbb{Z}/n)\cong Pic(X)[n].$$ I hope, this is of some use...

(The cohomology groups used above are the etale ones.)

1

I am not sure whether this is exactly what you want. But still, I see the following relation between $Pic$ and $\pi_1$:

Let $k$ be an algebraically closed field and $X/k$ a proper variety. Then there is an isomorphism $H^1(X, \mathbb{Z}/n)=Hom(\pi_1(X), \mathbb{Z}/n)$. Furthermore, from the long exact cohomology sequence associated to $$0\to \mathbb{Z}/n\to \mathbb{G}_m\to \mathbb{G}_m\to 0$$ we obtain an isomorphism $$H^1(X, \mathbb{Z}/n)\cong H^1(X, \mathbb{G}_m)[n].$$ Here we used that $k$ is algebraically closed and that $X$ is proper over $k$. By a generalization of Hilbert 90 we have $H^1(X, \mathbb{G}_m)=Pic(X)$. Hence, after all, we see that there is an isomorphism $$Hom(\pi_1(X), \mathbb{Z}/n)\cong Pic(X)[n].$$ I hope, this is of some use...