It is easy to come up with a metric space whose completion has strictly larger cardinality. Something like $\mathbb Q$ with completion $\mathbb R$ will do. Or more generally any subset of $\mathbb R^n$ has a countable dense subset we can use.
But for normed spaces over $\mathbb R$ this kind of funnybusiness is not allowed. $\mathbb Q$ is still dense in $\mathbb R$ but it is not a normed space over $\mathbb R$ like we want. The matter is complicated further by how all incomplete normed spaces bust have infinite dimension.
The standard example is the space $c_{00}$ of eventually-zero real sequences with completion equal to the space of sequences converging to zero. Here we use the norm $\|a\| = \sum_{i=1}^\infty |a_i|$. Unfortunately both these spaces have cardinality $\frak c$.
Can it happen that the completion is strictly larger than the original normed space?
I think so.
Let $|\Omega|\ge \frak c$ and let $c_{00}(\Omega)$ be the vector space of functions $\Omega \to \mathbb R$ that are zero on all but finitely many elements, equipped with the $\sup$ norm. To see the density is at least $|\Omega|$ consider the family of open balls of radius $1/2$ centred at the functions $\{c_x: x \in \Omega\}$ with entries $c_x(x) = 1$ and $c_x(y) = 0$ for $y \ne x$. These balls are pairwise disjoint so every dense subset needs at least one element per ball.
In the other direction, we claim the density is at most $|\Omega|$ because this is the cardinality of the space. To prove that, we construct a function $\mathcal P_{\text{fin}}(\Omega) \times \mathbb R^{\aleph_0} \to c_{00}(\Omega)$. The left-hand-side has cardinality $|\Omega| \times \frak c = |\Omega| $.
To construct the function, fix a total ordering on $\Omega$. Given $(P,a_1,a_2,\ldots) \in P_{\text{fin}}(\Omega) \times \mathbb R^{\aleph_0}$ write $P = \{p_1,p_2,\ldots p_n\}$ with $p_1 < p_2 < \ldots < p_n$ and assign it to the function $f$ with $ f(p_i) = a_i$ and $f(x) = 0$ for $x \notin P$.
This shows $c_{00}(\Omega)$ has density exactly $|\Omega|$. Since $c_{00}(\Omega)$ is dense in its completion, and the completion is a metric space, $c_{00}(\Omega)$ has the same density as the completion.
The completion is a Banach space, and therefore has cardinality exactly $|\Omega|^{\aleph_0}$ . One way to see this is to use how every Banach space is homomorphic to some Hilbert space $\ell^2(K)$ and then show directly $\ell^2(K)$ has density $|K|$ and cardinality $|K|^{\aleph_0}$.
In particular let $|\Omega|$ be a set have countable cofinality. Then Konig's theorem says $|\Omega|^{\aleph_0} > |\Omega|$. But the LHS is the cardinality of $c_{00}(\Omega)$ and the RHS is the cardinality of the completion.