This is rather a question for mathoverflow. The canonical map $\pi^* : \mathrm{Pic}(X)\to\mathrm{Pic}(X\times\mathbb A^1)$ is an isomorphism when $X$ is normal, and there are counterexamples with $X$ local integral of dimension $1$ and of course non-normal (with no isomorphism between $\mathrm{Pic}(X)$ and $\mathrm{Pic}(X\times\mathbb A^1)$).
First suppose $X$ is normal. As $\pi^*$ is injective, it is enough to consider affine open subsets of $X$. Then descend to finitely generated $\mathbb Z$-algebras. Taking integral closure and because $\mathbb Z$ is excellent, we are reduced to the case $X$ noetherian and normal. Now apply EGA IV.21.4.11, page 360. Note that the same arguments apply for any non-empty open subset of $\mathbb A^1_\mathbb Z$, e.g., $\mathbb G_m$, instead of $\mathbb A^1$.
Now let us see a counterexample with $X$ non-normal. I will take the first integral non-normal example which comes to my mind: $X=\mathrm{Spec}(R)$ where $R$ is the local ring $$R=(k[u,v]/(u^2+v^3))_{(u,v)}$$
over a field $k$ of characteristic zero. Let $K=\mathrm{Frac}(R)$. As $X$ is local, $\mathrm{Pic}(X)$ is trivial. Denote by $Y:=X\times \mathbb A^1$. We want to show $\mathrm{Pic}(Y)$ is non-trivial.
Consider the polynomial
$$f=1+vT^2\in R[T].$$
It is chosen in such a way that $f$ is not irreducible in $K[T]$ :
$$f=(1+tT)(1-tT)=(v+uT)(1/v+(v/u)T),\quad t:=-u/v\in K$$
but is $f$ irreducible (I don't say prime) in $R[T]$ (I don't use explicitly this property, just to explaine where comes this $f$). Note that $Y$ is covered by the two affine open subsets $D(f)$ and the generic fiber $Y_K$. Let $L$ be the invertible sheaf on $Y$ given by
$$L({D(f)})=(v+uT)R[T]_f, \quad L({Y_K})=K[T].$$
This is well defined because $(v+uT)$ is an invertible element of $O_Y(D(f)\cap Y_K)=K[T]_f$.
Admit for a moment that
$(R[T]_f)^{\star}=R^{\star}f^{\mathbb Z}$.
If $L$ is free, then there exist $\omega=af^{r}\in (R[T]_f)^\star$ and $\lambda\in (K[T])^\star=K^\star$ such that $(v+uT)\omega=\lambda$. This is impossible by comparing the degrees of both sides in $K[T]$. So $\mathrm{Pic}(Y)$ is non-trivial. The above fact on the units of $R[T]_f$ is (with the new $f$) easy to see. But I can post my solution if you want.
Over an algebraically closed field, see Qiaochu's remark. Over a number field, it is always a finitely generated abelian group. If $g=1$ and the base field is a given number field, then it is believed there exist elliptic curves of arbitrarily large rank, but this is not known (I believe the largest rank known over $\mathbf Q$ is 28, due to Elkies).
Best Answer
Assume the base field is $\mathbb{C}$.
If $n = 1$, $X \cong \mathbb{P}^1 \times \mathbb{P}^1$ (as Mohan said), so $Pic(X) \cong \mathbb{Z} \oplus \mathbb{Z}$ and $X$ can be embedded into $\mathbb{P}^3$.
If $n = 2$, $X$ is a del Pezzo surface of degree 2, it is isomorphic to $\mathbb{P}^2$ blown up at 7 points, $Pic(X) \cong \mathbb{Z}^8$, $X$ cannot be embedded into $\mathbb{P}^3$.
If $n = 3$, $X$ is a K3 surfaces. Generically, $Pic(X) \cong \mathbb{Z}$, but in the 19-dimensional moduli space of such covers there is a countable number of subspaces where the Picard rank jumps (up to 20).
If $n \ge 4$, $X$ is of general type, generically $Pic(X) \cong \mathbb{Z}$, but again it jumps on subvarieties of muduli space; this is governed by the Hodge theory.