Generally, given $T\colon \mathcal{D}(T) \to H_2$, where $H_1, H_2$ are Hilbert spaces, $\mathcal{D}(T)$ is a dense subspace of $H_1$, and $T$ is a linear operator, the adjoint of $T$ is defined on the subspace $\mathcal{D}(T^\ast) \subset H_2$ of elements $y$ such that there exists a $z\in H_1$ with $\langle Tx,y\rangle_2 = \langle x, z\rangle_1$ for all $x\in \mathcal{D}(T)$, then $T^\ast(y) = z$. In short,
$$\langle Tx,y\rangle_2 = \langle x, T^\ast y\rangle_1$$
for all $x\in \mathcal{D}(T),\, y \in \mathcal{D}(T^\ast)$. The denseness of $\mathcal{D}(T)$ ensures the well-definedness of $T^\ast$.
In the situation at hand, the codomain of $T$ is $\mathbb{K}$ (whether that's $\mathbb{R}$ or $\mathbb{C}$ doesn't matter), so there are two possibilities for $\mathcal{D}(T^\ast)$; it can be either $\{0\}$ or $\mathbb{K}$. If $\mathcal{D}(T^\ast) = \mathbb{K}$, then $T^\ast\colon \mathbb{K}\to L^2$ is continuous, and hence it has a continuous adjoint $T^{\ast\ast}\colon L^2 \to \mathbb{K}$. But we then have $T \subset T^{\ast\ast}$, so $T$ itself would be continuous. The given $T\colon f \mapsto f(0)$ is not continuous, hence $\mathcal{D}(T^\ast) = \{0\}$, i.e. $T^\ast$ is not densely defined.
The argument shows, with minor modifications, that a densely defined operator with finite-dimensional codomain has a densely defined adjoint if and only if it is continuous, since the only dense subspace of a finite-dimensional Hausdorff topological vector space is the entire space, and every linear operator $\mathbb{K}^n \to V$, where $V$ is a topological vector space, is continuous.
Here's a proof I learned from some notes of Richard Melrose. I just noticed another answer was posted while I was typing. This uses a different characterization of compactness so hopefully it is interesting for that reason.
First, I claim that a set $K\subset H$ of a Hilbert space is compact if and only if it is closed, bounded, and satisfies the equi-small tail condition with respect to any orthonormal basis. This means that given a basis $\{e_k\}_{k=1}^\infty$, for any $\varepsilon>0$ we can choose $N$ large enough such that for any element $u\in H$,
$$\sum_{k>N} |\langle u , e_k \rangle |^2 < \varepsilon.$$
The main point here is that this condition ensures sequential compactness. The proof is a standard "diagonalization" argument where you choose a bunch of subsequences and take the diagonal. This is done in detail on on page 77 of the notes I linked.
With this lemma in hand, the proof is straightforward. I repeat it from page 80 of those notes. Fix a compact operator $T$. By definition [this is where we use a certain characterization of compactness] the image of the unit ball $T(B(0,1))$ is compact. Then we have the tail condition that, for given $\varepsilon$, there exists $N$ such that
$$\sum_{k>N} |\langle Tu , e_k \rangle |^2 < \varepsilon.$$
for any $u$ such that $\| u \| < 1$.
We consider the finite rank operators
$$T_nu = \sum_{k\le n} \langle Tu , e_k \rangle e_k.$$
Now note the tail condition is exactly what we need to show $T_n \rightarrow T$ in norm. So we're done.
Best Answer
On $\ell^2$, define $A$ and $B$ by $(Ax)_n = -(Bx)_n = n^2 x_n$ for $n > 1$, $(A x)_1 = \sum_{n=1}^\infty n x_n$ and $(B x)_1 = 0$, with $D(A) = D(B) = \{x: \sum_{n =1}^\infty n^4 |x_n|^2 < \infty \}$. Then if I'm not mistaken $A$ and $B$ are closed but $A + B$ is not closable, e.g. (with $e_n$ the standard unit vectors) $\lim_{n \to \infty} e_n/n =0$ while $(A + B) e_n/n = e_1$.