McDuff states that she's assuming throughout that she's dealing with rational homology spheres only (i.e. finite $H_1$), which simplifies the answer to your question a great deal. Also, for the sake of simplicity I will only consider coefficients in $\mathbb{F}=\mathbb{F}_2$ (the field with two elements).

"Ignoring the basepoint" can be translated into algebra by saying "set $U=1$" (in all theories except the hat, where there's no $U$): this has the effect of working on the chain complex $\widehat{CF}$ but with a lot more differentials than $\widehat{\partial}$ normally counts.

Except that now we have a nice algebraic version of dealing with this, and one just needs to take the tensor product with $R = \mathbb{F}[U]/(U-1)\cong \mathbb{F}$, and when we're working over $\mathbb{F}$ the algebra is much kinder. In particular, this new version of $HF$ will be $HF^+(Y)\otimes R = HF^-(Y)\otimes R$.

Since, for rational homology spheres, $HF^\pm$ is simply one tower plus torsion *in each spin$^c$ structure*, and since tensoring with $R$ kills the torsion, what you're left with is simply a copy of $\mathbb{F}$ for each spin$^c$ structure.

For 3-manifolds with $b_1>0$ the situation is slightly more complicated, and I think it's now proven (at least using the isomorphism with Seiberg-Witten) that this new version only sees (co)homological data, but you have to take into account the triple cup product $H^1\otimes H^1\otimes H^1 \to H^3$. E.g., for $T^3$ I would expect that ignoring the basepoint you get $\mathbb{F}^6$, while for $\#^3(S^1\times S^2)$ (which has the same $H_1$ as $T^3$) you get $\mathbb{F}^8$.