Is it consistent with ZF that $V \to V^{\ast \ast}$ is always an isomorphism?

No, it’s not consistent.

Let $V=k^{(\omega)}$ be the vector space of finite sequences of elements of $k$. Then $V^*$ can be identified with the vector space $k^\omega$ of all sequences, and elements of the image of the natural map $V\to V^{**}$, considered as maps $k^\omega\to k$, are determined by their restriction to $k^{(\omega)}$.

So if $V\to V^{**}$ is an isomorphism, then, taking $W=k^\omega/k^{(\omega)}$, there are no nonzero linear maps $W\to k$, and hence $W^{**}=0$. But $W$ is nonzero.

So the map to the double dual must fail to be an isomorphism either for $V$ or for $W$.

Not an answer to your question but a variant which is perhaps more à propos given that you said you are teaching a rigorous undergrad analysis class.

Let $V=\oplus_{\mathbb{N}}\mathbb{R}$ be the space with countable basis which motivated your question. Another way to salvage the isomorphism $V\simeq V^{\ast\ast}$ (without giving up any cherished axiom) is to think of $V$ as equipped with the finest locally convex topology and consider duals as topological duals always given the strong topology. It is easy to see that $V^{\ast}\simeq \prod_{\mathbb{N}}\mathbb{R}$ with the product topology. Although perhaps counterintuitive, when one takes the proper (i.e., strong topological) dual of $\prod_{\mathbb{N}}\mathbb{R}$, then one gets back to $\oplus_{\mathbb{N}}\mathbb{R}$.