Theorem B of this paper implies that we can take any two nontrivial involution-free groups $A$ and $B$ and construct a
complete simple groups $D$ with a (diagrammatically) aspherical presentation $D=A*B/\langle\!\langle w_1, w_2,\dots\rangle\!\rangle$ (though such use of this theorem is killing a mosquito with a cannon).
Here, complete means naturally isomorphic to the automorphism group (i.e. centreless and without outer automorphisms). Now, suppose that $A$ and $B$ coincides with their commutator subgroups.
Acphericity implies that the centre of the free central extension $\widetilde D=A*B/[\langle\!\langle w_1, w_2,\dots\rangle\!\rangle,A*B]$ of $D$ is the free abelian group with the basis $\widetilde{w_1},\widetilde{w_2},\dots$ (see, e.g., Olshanskii's book, Section 34.4). Now, we can do whatever we want. For instance, we can take the quotient $G=\widetilde D/\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ and obtain a desired group $G$.
The natural map $\mbox{Aut}\,\widetilde D\to\mbox{Aut}\,G$ exists because the subgroup $\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ is characteristic in $\widetilde D$ (as this subgroup consists of $p$th powers of all central elements), and this map is injective because
$\widetilde D$ coincides with its commutator subgroup. So, there are no outer automorphisms of $G$.
Jul 6, 2015. More details added, as suggested by Mamuka.
(1)
$G$ (as well as $D$ and $\widetilde D$) is finitely generated if $A$ and $B$ are finitely generated.
(2)
Suppose that a group is a quotient of another group: $L=M/N$. Then
there is a natural map $f\colon \mbox{Aut}\,M\to \mbox{Aut}\,L$ if $N$ is characteristic;
$f(\mbox{Inn}\,M)=\mbox{Inn}\,L$ (i.e. $f$ sends inner automorphisms to inner automorphisms, and each inner automorphism of $L$ has at least one inner preimage);
$f$ is injective if $N$ is central and $M$ coincides with its commutator subgroup.
Now, take $M=G$ and $L=D$ (and $N=\langle\widetilde {w_1},\widetilde{w_2},\dots\rangle$). Then $\mbox{Aut}\,L=\mbox{Inn}\,L$ and, hence,
$\mbox{Aut}\,M=\mbox{Inn}\,M$ by (2), i.e.
all automorphisms of $M=G$ are inner.
Best Answer
A nice set of generators for the automorphism group of a finite abelian group is described by Garrett Birkhoff in his paper titled "Subgroups of abelian groups", Proc. London Math. Soc., s2-38(1):385-401, 1935 (MR).
Note that each finite abelian group is the product of its $p$-primary subgroups. An automorphism preserves the primary parts. So it is sufficient to consider $p$-abelian groups.
It is convenient to think of automorphisms of finite abelian groups as integer matrices. Expressing the group $A = \mathbf Z/p^{\lambda_1}\oplus \dotsb \oplus \mathbf Z/p^{\lambda_n}$ as a quotient of the free abelian group $\mathbf Z^n$, lift an automorphism $\phi$ of $A$ to an automorphism $\tilde\phi$ of $\mathbf Z^n$: $$ \begin{matrix} \mathbf Z^n & \xrightarrow{\tilde \phi} & \mathbf Z^n\\ \downarrow & &\downarrow\\ A & \xrightarrow{\phi} & A \end{matrix} $$ The matrix $(\phi_{ij})$ representing $\tilde\phi$ is an invertible integer matrix. As far as the automorphism $\phi$ is concerned, the its entries in the $i$th row are in $\mathbf Z/p^{\lambda_i}\mathbf Z$. Also $\phi_{ij}$ is divisible by $p^{\max(0, \lambda_j-\lambda_i)}$. With this matrix representation, it is easy to do calculations. For example, composition is matrix multiplication.
As for your query concerning cardinalities: if $\lambda_1>\lambda_2>\dotsb>\lambda_n$ and $$ A = (\mathbf Z/p^{\lambda_1}\mathbf Z)^{\oplus m_1}\oplus\dotsc \oplus (\mathbf Z/p^{\lambda_n}\mathbf Z)^{\oplus m_n}, $$ then it is possible to deduce from the above description of the automorphism group that $$ \lvert\mathrm{Aut}(A)\rvert = q^{\sum_{i,j}m_im_j\min(\lambda_i,\lambda_j)}\prod_{k=1}^n \prod_{l=1}^{m_k} (1 - q^{-l}). $$ The group $\prod_{k=1}^n GL_{m_k}(\mathbf Z/p\mathbf Z)$ is a quotient of $\mathrm{Aut}(A)$ by a $p$-group.