Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(\Bbb N)$ is precisely the set $E=\{ \alpha e_i:i\in\Bbb N, |\alpha|=1 \}$ where
$e_i=(0,0,\dots,0,1,0,0,\dots)$, consisting of $0$ for all but the $i^{\text{th}}$ coordinate.
Proof: If $e_i = (1-\lambda)x+\lambda y$ for some $x,y\in B_{l^1}$, then $(1-\lambda)x_i+\lambda y_i=1$ and $|x_i|,|y_i|\le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $\alpha e_i$.
On the other hand, any point $x\in B_{l^1}\backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED
Having known that, it is easy to see that for any $x\in \text{co}(E)$, $x$ can have only finitely many nonzero terms, so $\text{co}(E) \subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $\overline{\text{co}} ^*(E)=B_{l^1}$ so $\text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $\text{co}(E)=\text{co}(\text{Ext}(K))$ is not weak$^*$ compact.
Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $\overline{\text{co}}(E)=B_{l^1}$:
We clearly have
$$
B_{l^1} = \overline{\text{co}}(E) \subset \overline{\text{co}} ^*(E)
$$
since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion
$$
\overline{\text{co}} ^*(E) \subset B_{l^1}
$$
comes from the fact that $\text{co}(E) \subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that
$$\overline{\text{co}} ^*(E)=B_{l^1}.$$
In my opinion yes, since you know have a new topology, you should prove again that the new topology
leads to a topological vector space.
Your attempt in the comments sounds correct to me so far.
Of course, one should also show that the multiplication by a scalar is continuous.
Best Answer
Such an open neighborhood $V$ can only be found if $K$ is contained in some finite dimensional subspace of $X$.
To see why, suppose that $V$ exists. By definition of the weak* topology, there are vectors $x_1,x_2,\ldots , x_n\in X$, and $\varepsilon >0$, such that $$ U:= \{x'\in X': |\langle x_i, x'\rangle |<\varepsilon \}\subseteq V. $$
Defining $y_i=\displaystyle\frac{2x_i}{\varepsilon }$, notice that, for all $x'$ in $X'$, one has that $$ |\langle y_i, x'\rangle |\leq 1 \Rightarrow |\langle x_i, x'\rangle |\leq \frac\varepsilon 2 < \varepsilon . $$ This implies that $$ \{y_1,y_2,\ldots , y_n\}^\circ \subseteq U\subseteq V, $$ whence $$ K\subseteq V^\circ \subseteq \{y_1,y_2,\ldots , y_n\}^{\circ \circ }. $$ Since the latter is the the circled, weakly closed convex hull of $\{y_1,y_2,\ldots , y_n\}$, which in turn is contained in $\text{span}\{y_1,y_2,\ldots , y_n\}$, we are done.