With the constraint that the cardinality $\kappa$ of a Hamel basis shall satisfy $\kappa^{\aleph_0} = \kappa$, the answer is affirmative.
For an infinite set $S$, let us denote the family of countably infinite subsets of $S$ by $\mathfrak{P}_\omega(S)$. Then we look at the space
$$\ell^2(S) = \biggl\{ f\colon S \to \mathbb{K} : \sum_{s\in S} \lvert f(s)\rvert^2 < +\infty\biggr\},$$
where $\mathbb{K} \in \{\mathbb{R},\mathbb{C}\}$ is the desired scalar field. It is elementarily (but maybe tediously) verified that $\ell^2(S)$ is a Hilbert space with the norm
$$\lVert f\rVert_2 = \sqrt{\sum_{s\in S} \lvert f(s)\rvert^2}$$
if that is not yet known (e.g. from measure theory, since $\ell^2(S) = L^2(S,\mu)$ where $\mu$ is the counting measure on the $\sigma$-algebra $\mathfrak{P}(S)$ of all subsets of $S$). Further, the family $\{ e_s : s\in S\}$, where
$$e_s(t) = \begin{cases} 1 &, t = s\\ 0 &, t\neq s, \end{cases}$$
is a Hilbert basis of $\ell^2(S)$, so $\dim_{\text{Hilbert}} \ell^2(S) = \operatorname{card} S$. Note that by Parseval's theorem, every Hilbert space is isometrically isomorphic to some $\ell^2(S)$ (we can take any Hilbert basis as $S$), and - since the Hilbert dimension is well-defined - $\ell^2(S)$ and $\ell^2(T)$ are isometrically isomorphic if and only if $\operatorname{card} S = \operatorname{card} T$.
Since a family of non-negative real numbers can have a finite sum only if at most countably many members of the family are non-zero, we have
$$\ell^2(S) = \bigcup_{T\in \mathfrak{P}_\omega(S)} \iota_T\bigl(\ell^2(T)\bigr),$$
where $\iota_T \colon \ell^2(T) \hookrightarrow \ell^2(S)$ is the natural (isometric) embedding
$$\iota_T(f)(s) = \begin{cases} f(s) &, s\in T\\ 0 &, s\notin T.\end{cases}$$
For every $T\in \mathfrak{P}_\omega(S)$, we have $\ell^2(T) \cong \ell^2(\mathbb{N})$, so $\operatorname{card} \ell^2(T) = \mathfrak{c} = 2^{\aleph_0}$. It follows that
\begin{align}
\operatorname{card} \ell^2(S) &= \operatorname{card} \Biggl(\bigcup_{T\in \mathfrak{P}_\omega(S)} \iota_T\bigl(\ell^2(T)\bigr)\Biggr)\\
&\leqslant \sum_{T\in \mathfrak{P}_\omega (S)} \operatorname{card} \ell^2(T)\\
&= \mathfrak{c}\cdot \operatorname{card} \mathfrak{P}_\omega(S)\\
&= \mathfrak{c}\cdot (\operatorname{card} S)^{\aleph_0}\\
&= \max \bigl\{\mathfrak{c}, (\operatorname{card} S)^{\aleph_0}\bigr\}\\
&= (\operatorname{card} S)^{\aleph_0}.
\end{align}
On the other hand, for every $T\in \mathfrak{P}_\omega(S)$, we can choose an $f_T\in \ell^2(S)$ such that $\langle f_T, e_s\rangle \neq 0 \iff s\in T$, and of course $f_{T_1} \neq f_{T_2}$ if $T_1 \neq T_2$, so $T\mapsto f_T$ gives an injection $\mathfrak{P}_\omega(S) \hookrightarrow \ell^2(S)$, and therefore
$$\operatorname{card} \ell^2(S) \geqslant \operatorname{card} \mathfrak{P}_\omega(S) = (\operatorname{card} S)^{\aleph_0}.$$
Together, we have
$$\operatorname{card} \ell^2(S) = (\operatorname{card} S)^{\aleph_0},$$
whence
$$\dim_{\text{Hamel}} \ell^2(S) = (\operatorname{card} S)^{\aleph_0}.$$
Thus, if $\mathfrak{s} = \operatorname{card} S$ satisfies $\mathfrak{s}^{\aleph_0} = \mathfrak{s}$ (and only then [we're not talking about the finite-dimensional case]), we have
$$\dim_{\text{Hilbert}} \ell^2(S) = \dim_{\text{Hamel}} \ell^2(S).$$
Now the construction is easy: If $V$ is a $\mathbb{K}$-vector space with a Hamel basis of cardinality $\kappa$ where $\kappa^{\aleph_0} = \kappa$, then we let $W = \ell^2(\kappa)$. By the above, $W$ is a Hilbert space whose algebraic dimension is $\kappa^{\aleph_0} = \kappa$, thus we can find an isomorphism $\varphi \colon V \to W$. Define
$$\langle u,v\rangle_V := \langle \varphi(u), \varphi(v)\rangle_W$$
to obtain an inner product on $V$ that makes $V$ a Hilbert space [since $\varphi$ is an isometry by definition of $\langle\,\cdot\,,\,\cdot\,\rangle_V$].
