I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list.

If you're willing to accept that it is consistent that $\aleph_1 < 2^{\aleph_0}$, you can get a relatively small example of an impossible Hamel dimension for a Banach space. The space $\ell^2$ contains a linearly independent set of cardinality $2^{\aleph_0}$, specifically $(n^{-\alpha})_{n \in \mathbb{N}}$ where $\alpha$ ranges over $[0,1]$. So the Hamel dimension of $\ell^2$ is at least $2^{\aleph_0}$, and can't be more because the cardinality of $\ell^2$ is $2^{\aleph_0}$. Now, for every infinite-dimensional Banach space $E$ you can build an injective bounded linear map $\ell^2 \rightarrow E$ (this is one of those constructions where it is simpler to just try it yourself than to follow someone else's way of doing it). So $\dim(E) \geq 2^{\aleph_0} > \aleph_1$ and there is no Banach space of Hamel dimension $\aleph_1$.

In general, in Lemma 2 of

Kruse, Arthur H., Badly incomplete normed linear spaces, Math. Z. 83, 314-320 (1964). ZBL0117.08201.

Kruse showed that for a Banach space $E$, $\dim(E)^{\aleph_0} = \dim(E)$. By König's theorem, if $\kappa$ is uncountable and the union of countably many strictly smaller sets, then $\kappa^{\aleph_0} > \kappa$. So, unconditionally, there is no Banach space of Hamel dimension $\aleph_\omega = \bigcup_{n=0}^\infty \aleph_n$ nor of Hamel dimension $\beth_\omega = \bigcup \{ 2^{\aleph_0}, 2^{2^{\aleph_0}}, 2^{2^{2^{\aleph_0}}}, \ldots \}$.

(It seems that Kruse's lemma does not require the axiom of replacement. If this is so, then we cannot find an unconditional counterexample without using the axiom of replacement because in the $V_{\omega + \omega}$ inside $L$ we have that $X^{\omega} \cong X$ for every uncountable set.)

I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list.

If you're willing to accept that it is consistent that $\aleph_1 < 2^{\aleph_0}$, you can get a relatively small example of an impossible uncountable Hamel dimension for a Banach space. The space $\ell^2$ contains a linearly independent set of cardinality $2^{\aleph_0}$, specifically $(n^{-\alpha})_{n \in \mathbb{N}}$ where $\alpha$ ranges over $[0,1]$. So the Hamel dimension of $\ell^2$ is at least $2^{\aleph_0}$, and can't be more because the cardinality of $\ell^2$ is $2^{\aleph_0}$. Now, for every infinite-dimensional Banach space $E$ you can build an injective bounded linear map $\ell^2 \rightarrow E$ (this is one of those constructions where it is simpler to just try it yourself than to follow someone else's way of doing it). So $\dim(E) \geq 2^{\aleph_0} > \aleph_1$ and there is no Banach space of Hamel dimension $\aleph_1$.

In general, in Lemma 2 of

Kruse, Arthur H., Badly incomplete normed linear spaces, Math. Z. 83, 314-320 (1964). ZBL0117.08201.

Kruse showed that for a Banach space $E$, $\dim(E)^{\aleph_0} = \dim(E)$. By König's theorem, if $\kappa$ is uncountable and the union of countably many strictly smaller sets, then $\kappa^{\aleph_0} > \kappa$. So, unconditionally, there is no Banach space of Hamel dimension $\aleph_\omega = \bigcup_{n=0}^\infty \aleph_n$ nor of Hamel dimension $\beth_\omega = \bigcup \{ 2^{\aleph_0}, 2^{2^{\aleph_0}}, 2^{2^{2^{\aleph_0}}}, \ldots \}$.

(It seems that Kruse's lemma does not require the axiom of replacement. If this is so, then we cannot find an unconditional counterexample without using the axiom of replacement because in the $V_{\omega + \omega}$ inside $L$ we have that $X^{\omega} \cong X$ for every uncountable set.)

I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list.

If you're willing to accept that it is consistent that $\aleph_1 < 2^{\aleph_0}$, you can get a relatively small example of an impossible Hamel dimension for a Banach space. The space $\ell^2$ contains a linearly independent set of cardinality $2^{\aleph_0}$, specifically $(n^{-\alpha})_{n \in \mathbb{N}}$ where $\alpha$ ranges over $[0,1]$. So the Hamel dimension of $\ell^2$ is at least $2^{\aleph_0}$, and can't be more because the cardinality of $\ell^2$ is $2^{\aleph_0}$. Now, for every infinite-dimensional Banach space $E$ you can build an injective bounded linear map $\ell^2 \rightarrow E$ (this is one of those constructions where it is simpler to just try it yourself than to follow someone else's way of doing it). So $\dim(E) \geq 2^{\aleph_0} > \aleph_1$ and there is no Banach space of Hamel dimension $\aleph_1$.

In general, in Lemma 2 of

Kruse, Arthur H., Badly incomplete normed linear spaces, Math. Z. 83, 314-320 (1964). ZBL0117.08201.

Kruse showed that for a Banach space $E$, $\dim(E)^{\aleph_0} = \dim(E)$. By König's theorem, if $\kappa$ is uncountable and the union of countably many strictly smaller sets, then $\kappa^{\aleph_0} > \kappa$. So, unconditionally, there is no Banach space of Hamel dimension $\aleph_\omega = \bigcup_{n=0}^\infty \aleph_n$ nor of Hamel dimension $\beth_\omega = \bigcup \{ 2^{\aleph_0}, 2^{2^{\aleph_0}}, 2^{2^{2^{\aleph_0}}}, \ldots \}$.

(It seems that Kruse's lemma does not require the axiom of replacement. If this is so, then we cannot find an unconditional counterexample without using the axiom of replacement because in the $V_{\omega + \omega}$ inside $L$ we have that $X^{\omega} \cong X$ for every uncountable set.)

I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list.

If you're willing to accept that it is consistent that $\aleph_1 < 2^{\aleph_0}$, you can get a relatively small example of an impossible Hamel dimension for a Banach space. The space $\ell^2$ contains a linearly independent set of cardinality $2^{\aleph_0}$, specifically $(n^{-\alpha})_{n \in \mathbb{N}}$ where $\alpha$ ranges over $[0,1]$. So the Hamel dimension of $\ell^2$ is at least $2^{\aleph_0}$, and can't be more because the cardinality of $\ell^2$ is $2^{\aleph_0}$. Now, for every infinite-dimensional Banach space $E$ you can build an injective bounded linear map $\ell^2 \rightarrow E$ (this is one of those constructions where it is simpler to just try it yourself than to follow someone else's way of doing it). So $\dim(E) \geq 2^{\aleph_0} > \aleph_1$ and there is no Banach space of Hamel dimension $\aleph_1$.

In general, in Lemma 2 of

Kruse, Arthur H., Badly incomplete normed linear spaces, Math. Z. 83, 314-320 (1964). ZBL0117.08201.

Kruse showed that for a Banach space $E$, $\dim(E)^{\aleph_0} = \dim(E)$. By König's theorem, if $\kappa$ is uncountable and the union of countably many strictly smaller sets, then $\kappa^{\aleph_0} > \kappa$. So, unconditionally, there is no Banach space of Hamel dimension $\aleph_\omega = \bigcup_{n=0}^\infty \aleph_n$ nor of Hamel dimension $\beth_\omega = \bigcup \{ 2^{\aleph_0}, 2^{2^{\aleph_0}}, 2^{2^{2^{\aleph_0}}}, \ldots \}$.

(It seems that Kruse's lemma does not require the axiom of replacement. If this is so, then we cannot find an unconditional counterexample without using the axiom of replacement because in the $V_{\omega + \omega}$ inside $L$ we have that $X^{\omega} \cong X$ for every uncountable set.)