We use the Seifert Van-Kampen Theorem to calculate the fundamental group of a connected graph. This is Hatcher Problem 1.2.5:
It is a fact in graph theory that any connected graph $X$ contains a maximal tree $M$, namely a contractible graph that contains all the vertices of $X$. Now if the maximal tree $M = X$, then we are done because for any $x_0 \in M$, $\pi_1(M,x_0) = \pi_1(X,x_0) = 0$ that is trivially free. Now suppose $M \neq X$. Then there is an edge $e_i$ of $X$ not in $M$. Observe that for each edge $e_i$ we get a loop going in $M \cup e_i$ about some point $x_0 \in M$. Now fix out basepoint $x_0$ to be in $M$ and suppose that the edges not in $M$ are $e_1,\ldots,e_n$. Then it is clear that
$$X = \bigcup_{i=1}^n \left(M \cup e_i\right).$$
The intersection of any two $M \cup e_i$ and $M \cup e_j$ contains at least $M$ and is path connected, so is the triple intersection of any 3 of these guys by the assumption that $X$ is a connected graph. So for any $x_0 \in M$, the Seifert-Van Kampen Theorem now tells us that
$$\pi_1(X,x_0) \cong \pi_1(M \cup e_1,x_0) \ast \ldots \ast \pi_1(M \cup e_n,x_0)/N$$
where $N$ is the subgroup generated by words of the form $l_{ij}(w)l_{ji}(w)^{-1}$, where $l_{ij}$ is the inclusion map from $\pi_1((M\cup e_i) \cap (M \cup e_j),x_0) = \pi_1(M \cup (e_i \cap e_j),x_0)$. Now observe that if $i \neq j$ then $M \cup (e_i \cap e_j) = M$ and since $\pi_1(M,x_0) = 0$ we conclude that any loop $w \in \pi_1(M \cup (e_i \cap e_j),x_0)$ in here is trivial. If $i = j$, $l_{ij}$ is just the identity so that our generators for $N$ are just
$$l_{ij}(w)l_{ji}(w)^{-1} = ww^{-1} = 1$$
completing our claim that $N$ was trivial. Now for each $i$, we have that $\pi_1(M\cup e_i,x_0)$ is generated by a loop that starts at $x_0$ and goes around the bounded complementary region formed by $M$ and $e_i$ and back to $x_0$ through the maximal tree. Such a path back to $M$ does not go through any other edge $e_j$ for $j$ different from $i$. It follows that $\pi_1(X,x_0)$ is a free group with basis elements consisting of loops about $x_0 \in M$ as described in the line before.
Under certain conditions they do actually mean the existence of neighborhoods, containing the common point $x_0$ of the wedge, which do retract on $x_0$.
With this condition we are able to choose suitable open path-connected sets $A_i$ to compute the desired fundamental group.
For instance if you take the $n$ wedge of circles $S^1$ at $x_0$ you can compute its fundamental group via Van-Kampen. What is important here is that we have open neighborhoods $N_i$ containing $x_0$ for each circle $S^1$ of the wedge (just imagine a little open arc on $S^1$ which contains the center $x_0$ of the wedge) which do actually retract on $x_0$ (just shrink from both sides until you reach the center $x_0$ of the wedge).
Now you can indeed define $A_i$ to be the wedge of all $N_j$ for $j \neq i$ and wedge $S^1$ i.e. $$A_i=\left(\bigvee_{i \neq j}N_j\right) \vee S^1.$$
If you now take two different $A_i$'s, i.e. $A_i$ and $A_j$ and look at the intersection $A_i \cap A_j$ you'll see that it's simply the wedge of all $n$ neighboorhoods $N_i$ of $x_0$ and since all of them retract to $x_0$ we get $$\pi_1\left( \bigvee_{i=1}^nN_i\right) \cong \pi_1(\{x_0\})=0.$$
By a similiar argument using your retracts you can compute
$$\pi_1(A_i) \cong \pi_1(S^1) \cong \mathbb Z.$$
Now by applying Van Kampen you get that the fundamental group of your wedge sum is free abelian of rank $n$.
Best Answer
A counterexample to the conjecture is given on p.44 (in my edition) of Hatcher's book on algebraic topology. You really do need the 3-fold intersection condition.
However there is a method of dealing with this which I published in 1967, namely to use many base points and so use the fundamental groupoid $\pi_1(B,S)$ on a set $S$ chosen according to the geometry of the situation: there is discussion of this idea in this mathoverflow entry. The idea was thought of in order to get a theorem which would compute at the same time also the fundamental group of the circle, a rather important example in topology, as well as a myriad of other cases. Why not?
The most general theorem of this type seems to be the following, from this paper, of 1984. Notice that for convenience we write also $\pi_1(U,S)$ for $\pi_1(B, U \cap S)$.
Theorem Let $(B_\lambda)_{\lambda\in \Lambda}$ be a family of subspaces of $B$ such that the interiors of the sets $B_\lambda$ $(\lambda\in \Lambda)$ cover $B$, and let $S$ be a subset of $B$. Suppose $S$ meets each path-component of each one-fold, two-fold, and each three-fold intersection of distinct members of the family $(B_\lambda)_{\lambda\in \Lambda}$. Then there is a coequalizer diagram in the category of groupoids: $$\bigsqcup_{\lambda,\mu\in\Lambda}\pi_1(U_\lambda\cap U_\mu, S)\rightrightarrows^\alpha_\beta \bigsqcup _{\lambda\in \Lambda}\pi_1(U_ \lambda,S)\to^\gamma\pi_1(B,S) $$ in which $\bigsqcup$ stands for the coproduct (disjoint union) in the category of groupoids, and $\alpha$, $\beta$, and $\gamma$ are determined by the inclusion maps $U_\lambda\cap U_\mu\to U_\lambda$, $U_\lambda\cap U_\mu\to U_\mu$, and $U_\lambda\to B$, respectively.
The proof involves verifying the universal property of a coequaliser, and so does not involve knowing how to construct coequalisers of groupoids. However this is discussed in the 1971 book by Philip Higgins "Categories and Groupoids" referred to in the mathoverflow discussion. I do not attempt here to find the "groupoid error term" in the case the 3-fold condition is not satisfied.
Grothendieck has supported the idea of "many base points" in his 1984 "Esquisse d'un programme" Section 2. Part of his comment is, in translation, "... people still obstinately persist, when calculating with fundamental groups, in fixing a single base point,,, ".