Let $M$ be the $4$-manifold $D^4\cup2\text{-handle}\cup\ldots\cup2\text{-handle}$, where the attachment of the handles is specfied by an oriented framed link $L=L_1\cup\ldots\cup L_n\subseteq S^3$. By Lickorish's theorem any smooth closed connected orientable 3-manifold can be obtained in this way as $\partial M$. For example, if $L$ consists of only one unknot with framing $n$, then $\partial M$ is the lens space $L(1,n)$. Thus in general, $H_1\partial M$ can be any finitely generated abelian group.
Does $\partial M$ obtain a handle decomposition from the one on $M$? I know that $\partial M$ is obtained as $S^3$ with $n$ $1$-surgeries. I would like to show that $H_1\partial M\cong \mathbb{Z}^n/\mathrm{Im}A$, where $A\in\mathbb{Z}^{n\times n}$ has on the diagonal the framings of $L_1,\ldots,L_n$ and at $ij$-th place the linking number of $L_i,L_j$.
There's a theorem in Ranicki's Algebraic and Geometric Surgery, p. 55, Proposition 4.19, describitng homology after surgery, burt it doesn't tell me much.
The exact sequence of a pair $H_2\partial M\rightarrow H_2M\rightarrow H_2(M,\partial M) \rightarrow H_1\partial M \rightarrow H_1M \rightarrow H_1(M,\partial M)$ with $H_1M=0=H_3M$ and $H_2M=\mathbb{Z}^n$ doesn't tell me much.
It would be desirable if the handle decomposition on $M$ which is uniquely specified by the link $L$ would induce a handle decomposition on $\partial M$, which would also be specified by $L$. If $\partial M$ had only $1$ $0$-handle and $n$ $1$-handles, then $\partial_1:\mathbb{Z}^{1\text{-handles}}\rightarrow\mathbb{Z}^{0\text{-handles}}$ would be the zero map, so $H_1\partial M=\mathbb{Z}^n/\mathrm{Im}\partial_2$ and then wishfully $\partial_2=A$.
So, how does the link $L$ induce a handle decomposition on $\partial M$?