Boundedness of a continuous linear functional on a topological vector space
The precise answer is as follows (according to Theorem 3.1 (f) in Conway's A Course in Functional Analysis):
Let $ X $ be a locally convex topological vector space. Suppose that $ \mathcal{P} = \{ \| \cdot \|_{i} \}_{i \in I} $ is a family of seminorms that defines the topology on $ X $. Then $ \Lambda $ is a continuous linear functional on $ X $ if and only if there exist $ i_{1},\ldots,i_{n} \in I $ and positive real numbers $ \lambda_{1},\ldots,\lambda_{n} $ such that $$ \forall x \in X: \quad |\Lambda(x)| \leq \sum_{k=1}^{n} \lambda_{k} \| x \|_{i_{k}}. $$
Addendum
As it seems that Conway does not provide a proof of the quoted theorem, I shall provide my own proof, as the OP has requested to see one.
There are many definitions of continuity at a point, so I shall pick the one that best suits my needs for the proof.
Definition Let $ X $ and $ Y $ be topological spaces. A function $ f: X \rightarrow Y $ is said to be continuous at $ x $ if and only if for all neighborhoods $ V $ of $ f(x) $, there exists a neighborhood $ U $ of $ x $ such that $ f[U] \subseteq V $.
We shall also use the fact that the continuity of $ \Lambda: X \rightarrow \mathbb{R} $ is equivalent to its continuity at the point $ 0_{X} $.
Let us first establish the ($ \Rightarrow $)-direction of the theorem. As (i) $ \Lambda(0_{X}) = 0 $ and (ii) $ \overline{D}(0;1) $ is a neighborhood of $ 0 $, by the given definition, there exists a neighborhood $ U $ of $ 0_{X} $ such that $ \Lambda[U] \subseteq \overline{D}(0;1) $. Without loss of generality, we may assume that $ U $ is a basic open neighborhood of the form $$ \{ x \in X \,|\, (\forall k \in \{ 1,\ldots,n \})(\| x \|_{i_{k}} < 2 \epsilon) \}, $$ where $ \epsilon > 0 $. Now, for each $ x \in X $, define $ \displaystyle M_{x} \stackrel{\text{def}}{=} \max_{1 \leq k \leq n} \| x \|_{i_{k}} $.
Claim: $ |\Lambda(x)| \leq \dfrac{M_{x}}{\epsilon} $ for all $ x \in X $.
Proof of the claim Let $ x \in X $. We shall consider two cases: (i) $ M_{x} = 0 $ and (ii) $ M_{x} > 0 $.
In Case (i), it is necessarily true that $ \Lambda(x) = 0 $. Suppose otherwise, i.e., $ |\Lambda(x)| = r > 0 $. Then for sufficiently large $ N \in \mathbb{N} $, we have $ |\Lambda(N \cdot x)| = Nr > 1 $. However, $ \| N \cdot x \|_{i_{k}} = N \| x \|_{i_{k}} = 0 $ for all $ k \in \{ 1,\ldots,n \} $, so $ N \cdot x \in U $. We have thus contradicted the earlier statement that $ \Lambda[U] \subseteq \overline{D}(0;1) $. Therefore, we must have $ \Lambda(x) = 0 $, in which case, the inequality $ |\Lambda(x)| \leq \dfrac{M_{x}}{\epsilon} $ automatically holds.
In Case (ii), we have $ \dfrac{\epsilon}{M_{x}} \cdot x \in U $, so $ \left| \Lambda \left( \dfrac{\epsilon}{M_{x}} \cdot x \right) \right| \leq 1 $. It follows immediately that $ |\Lambda(x)| \leq \dfrac{M_{x}}{\epsilon} $.
The claim is now established. //
Using the claim, we obtain $$ |\Lambda(x)| \leq \sum_{k=1}^{n} \frac{1}{\epsilon} \| x \|_{i_{k}}, $$ since $$ \sum_{k=1}^{n} \frac{1}{\epsilon} \| x \|_{i_{k}} = \frac{1}{\epsilon} \sum_{k=1}^{n} \| x \|_{i_{k}} \geq \frac{1}{\epsilon} \cdot M_{x}. $$
For the ($ \Leftarrow $)-direction, the argument is much easier. For any $ \epsilon > 0 $, if you take $ U $ to be the open set of $ X $ defined by $$ \left\{ x \in X \,\Bigg|\, (\forall k \in \{ 1,\ldots,n \}) \left( \| x \|_{i_{k}} < \frac{\epsilon}{n \cdot \max(\lambda_{1},\ldots,\lambda_{n})} \right) \right\}, $$ then you immediately obtain $ \Lambda[U] \subseteq D(0;\epsilon) $. Therefore, as $ \epsilon $ is arbitrary, $ \Lambda $ is continuous at $ 0_{X} $, hence continuous everywhere.
The proof of the theorem is now complete. ////