I believe the question is: "Does there exist a constant $A$ such that there exists infinitely many number fields with regulator less then $A$?" Ignore this response if that's incorrect.

I think the answer is no. The place to go for questions like this are the work of Zimmert and the survey paper of Odlyzko (which, as it turns out, would've pointed you to Zimmert anyway). You'll have to dig into the references to make sure I haven't misinterpreted a convention or notation, but I believe a proof goes as follows:

In Odlyzko's "Bounds for disciminants...", he cites Zimmert as giving a lower bound (equation 5.5) for the regulator of a number field which grows exponentially in the degree of the number field. So if the answer to your question were yes, your infinitely many number fields would have to be of bounded degree. But for number fields of a fixed degree, Theorem C in Friedman's "Analytic Formula for the Regulator of a Number Field" (attributed to Silverman, from an interesting-looking paper that I'd never heard of before) gives a lower bound for the regulator in terms of the discriminant D. This lower bound goes to infinity with D under the condition that all proper subfields of your number field have a strictly smaller unit rank, which is true for totally real number fields. The nail in the coffin is then that there are only finitely many number fields with discriminant under any given bound, and so only finitely many totally real number fields with regulator under any given bound.

Note that you can't cite the Silverman result right off the bat since the lower bound appears to me to go to 0 as $D$ and $n$ (discriminant and degree) simultaneously go to infinity.

Hope that helps.

I believe the question is: "Does there exist a constant $A$ such that there exists infinitely many totally real number fields with regulator less then $A$?" Ignore this response if that's incorrect.

I think the answer is no. The place to go for questions like this are the work of Zimmert and the survey paper of Odlyzko (which, as it turns out, would've pointed you to Zimmert anyway). You'll have to dig into the references to make sure I haven't misinterpreted a convention or notation, but I believe a proof goes as follows:

In Odlyzko's "Bounds for disciminants...", he cites Zimmert as giving a lower bound (equation 5.5) for the regulator of a number field which grows exponentially in the degree of the number field. So if the answer to your question were yes, your infinitely many number fields would have to be of bounded degree. But for number fields of a fixed degree, Theorem C in Friedman's "Analytic Formula for the Regulator of a Number Field" (attributed to Silverman, from an interesting-looking paper that I'd never heard of before) gives a lower bound for the regulator in terms of the discriminant D. This lower bound goes to infinity with D under the condition that all proper subfields of your number field have a strictly smaller unit rank, which is true for totally real number fields. The nail in the coffin is then that there are only finitely many number fields with discriminant under any given bound, and so only finitely many totally real number fields with regulator under any given bound.

Note that you can't cite the Silverman result right off the bat since the lower bound appears to me to go to 0 as $D$ and $n$ (discriminant and degree) simultaneously go to infinity.

Hope that helps.