If $(X, \|\cdot \|_X)$ is a normed vector space over $\mathbb R$, then the closed unit ball of $X$ is given by
$$B(X)=\{x\in X: \|x \|_X \le 1\}.$$
If $X^{*}$ is the set of all bounded linear functionals on the normed space $(X, \|\cdot \|_X)$ then the closed unit ball of the normed space $(X^*, \|\cdot \|_{op})$ is given by
$$B(X^*)=\{f\in X^*:\|f \|_{op}\le 1\},$$
where
$$\|f \|_{op}=\sup\left\{\frac{|f(x)|}{\|x \|_X}: x\in X \mbox{ with }x \neq 0 \right\}.$$
Suppose now that $X$ is a locally convex (Hausdorff) topological vector space with $X^*$ as its topological dual space. How can we describe the closed unit balls of $X^*$?
For a normed spaces we also have $$ \Vert f\Vert_{op}=\sup\left\{|f(x)|:x\in\operatorname{Ball}_X(0,1)\right\} $$ hence $$ \operatorname{Ball}_{X^*}(0,1)=\{f\in X^*:x\in\operatorname{Ball}_X(0,1)\implies |f(x)|\leq 1\} $$
For general Hausdorff locally convex spaces there is no single norm describing its topology. But there several seminorms which all together saves the day. So you can not define unit ball of $X$, you can not talk about norms of functionals and as the consequence you can not talk about unit ball in $X^*$. But there is very close notion of polar set defined as $$ V'=\{f\in X^*:x\in V\implies |f(x)|\leq 1\} $$ which is often can be suitable substitute for unit ball in dual space if you choose $V$ as neighbourhood of $0\in X$.