_{Allow me to steal my answer to a question from math.SE on the same topic. With minor changes}

In their paper, Howard and Tachtsis discuss these sort of questions. The paper was published rather recently, and I suspect that there hasn't been any significant progress since then.

Paul Howard and Eleftherios Tachtsis, On vector spaces over specific fields without choice, MLQ Math. Log. Q. 59 (2013), no. 3, 128--146.

In general, it seems that the axiom of regularity might play an important part in such proof. Which means that there might be more than just the structure of the vector spaces involved (although this might be mitigated by going at it as Blass did, by showing that you can prove $\sf MC$ rather than $\sf AC$).

If we slightly extend the result to "Every spanning set includes a basis", which in particular means that every vector space has a basis, then Keremedis showed in his paper cited below, that over $\Bbb Q$ if every generating set includes a basis, then the axiom of choice holds.

Kyriakos Keremedis, Extending independent sets to bases and the axiom of choice, Math. Logic Quart. 44 (1998), no. 1, 92--98.

I don't know the proof, but it might be possible to extend it to $\Bbb R$ or $\Bbb C$.

All in all, it seems that the questions you particularly address are still very much open, and perhaps new techniques are needed before we can find answers.

_{Allow me to steal my answer to a question from math.SE on the same topic. With minor changes}

In their paper, Howard and Tachtsis discuss these sort of questions. The paper was published rather recently, and I suspect that there hasn't been any significant progress since then.

Paul Howard and Eleftherios Tachtsis, On vector spaces over specific fields without choice, MLQ Math. Log. Q. 59 (2013), no. 3, 128--146.

In general, it seems that the axiom of regularity might play an important part in such proof. Which means that there might be more than just the structure of the vector spaces involved (although this might be mitigated by going at it as Blass did, by showing that you can prove $\sf MC$ rather than $\sf AC$).

If we slightly extend the result to "Every spanning set includes a basis", which in particular means that every vector space has a basis, then Keremedis showed in his paper cited below, that over $\Bbb Q$ if every generating set includes a basis, then the axiom of choice holds.

Kyriakos Keremedis, Extending independent sets to bases and the axiom of choice, Math. Logic Quart. 44 (1998), no. 1, 92--98.

I don't know the proof, but it might be possible to extend it to $\Bbb R$ or $\Bbb C$.

All in all, it seems that the questions you particularly address are still very much open, and perhaps new techniques are needed before we can find answers.