There is no such ring.

To lessen my typing, let me introduce some abbreviations.
C = commutative,
L = local,
D = dual,
K = Krull dimension greater than $0$,
$A(I)$ := $\textrm{Ann}(I)$ for $I\lhd R$,
N = $\textrm{Nil}(R)$.

**Theorem.** If $R$ is a CLDK ring, then $A(N)\subseteq N$.

The proof requires the following

**Lemma.** If $R$ is a CLD ring that is not a field
and $\mathfrak p\lhd R$ is a
prime ideal, then $A(\mathfrak p)\subseteq \mathfrak p$.

**Proof of Lemma.**
If $\mathfrak p$ is the maximal ideal, then $A(\mathfrak p)$
is a minimal ideal. Since $R$ is not a field,
this yields $R\neq A(\mathfrak p)$,
so $A(\mathfrak p)\subseteq \mathfrak p$ by locallness of $R$.
For the rest of the proof we consider only the case where
$\mathfrak p$ is nonmaximal.

As is well known,
if $\mathfrak p$ is a prime ideal, then it is $\cap$-irreducible.
(I.e., it is finitely meet irreducible.)
Reason: if $\mathfrak p = I\cap J$ and $\mathfrak p < I, J$,
then we contradict primeness by $IJ\subseteq I\cap J = \mathfrak p$.

**Claim.** If $\mathfrak p$ is nonmaximal, then it is *not*
$\bigcap$-irreducible. (I.e., $\mathfrak p$ is not infinitely meet irreducible.) Said another way, $\mathfrak p$ *will* equal the
complete intersection of all ideals that are properly above $\mathfrak p$.

**Proof of Claim.** Else $\mathfrak p^*:=\bigcap_{\mathfrak p < I} I$ is the smallest ideal
strictly above $\mathfrak p$. Then $\mathfrak p^*/\mathfrak p$ is the smallest nonzero ideal
in the domain $R/\mathfrak p$. Notice that this domain is not a field, since
$\mathfrak p$ was nonmaximal, so by its minimality $\mathfrak p^*/\mathfrak p$ is a proper ideal
of $R/\mathfrak p$.

$R/\mathfrak p$
is a domain and $0 < \mathfrak p^*/\mathfrak p\cdot \mathfrak p^*/\mathfrak p\leq \mathfrak p^*/\mathfrak p$,
forcing $\mathfrak p^*/\mathfrak p = (\mathfrak p^*/\mathfrak p)^2$. Now minimal idempotent ideals in commutative rings, like
$\mathfrak p^*/\mathfrak p$, are generated by idempotents, meaning
$\mathfrak p^*/\mathfrak p = (e)$ for some $e\in R$ satisfying $e^2=e$. The
element $e$ cannot be zero,
because $\mathfrak p^*/\mathfrak p$
is not zero, and it cannot be $1$, since $\mathfrak p^*/\mathfrak p$
is not $R/\mathfrak p$. Thus the domain $R/\mathfrak p$ has a proper idempotent $e$, which is absurd.
($e(1-e) = 0\neq e, (1-e)$ contradicts the definition of a domain.). This proves the claim. \\

Now back to the proof of the lemma. By the claim, if
$\mathfrak p$ is not maximal, then it equals the
intersection of all the ideals that properly contain it.
Applying the lattice anti-isomorphism $I\mapsto A(I)$, which must preserve
the complete lattice operations of the ideal lattice of $R$,
we obtain that $A(\mathfrak p)$ is the join $\bigvee_{\mathfrak p < I} A(I)$ of all ideals of the
form $A(I)$ where $\mathfrak p < I$. For any such $I$ we have
$$
I\cdot A(I) = 0\subseteq \mathfrak p,
$$
and $I \not\subseteq \mathfrak p$, so $A(I)\subseteq \mathfrak p$.
Hence the join $\bigvee_{\mathfrak p < I} A(I) = A(\mathfrak p)$
is also contained
in $\mathfrak p$. \\

Now

**Proof of Theorem.**
Suppose $R$ is a CLDK ring and that $\mathfrak p$ is a minimal prime
of $R$. If $\mathfrak q$ is a different minimal prime, then
$$
\mathfrak p \cdot A(\mathfrak p) = 0 \subseteq \mathfrak q,
$$
so either $\mathfrak p\subseteq \mathfrak q$ or
$A(\mathfrak p)\subseteq \mathfrak q$. The former cannot happen,
since $\mathfrak q$ is assumed to be minimal and different from $\mathfrak p$,
so we must have the latter. To reiterate: if $\mathfrak p$ is a minimal prime of $R$, then $A(\mathfrak p)$ is contained in every minimal prime different from $\mathfrak p$.

By the lemma, $A(\mathfrak p)\subseteq \mathfrak p$ as well,
so $A(\mathfrak p)$ is contained in all of the minimal primes. In other
words, if $\mathfrak p$ is a minimal prime, then
$$
A(\mathfrak p)\subseteq \bigcap_{\mathfrak q \textrm{min}} \mathfrak q
= N.
$$
Dualizing this yields $A(N)\subseteq \mathfrak p$ for every minimal prime $\mathfrak p$. It follows from this that
$$
A(N)\subseteq \bigcap_{\mathfrak p \textrm{min}} \mathfrak p
= N.
$$
\\

Quasi-Frobenius ringsit is example 6.6 on page 133 ("Clark example"). It's also $P$-injective, simple-injective but not self-injective. I thought for some time about generalizing it to higher Krull dimension. I was thinking that it might work for any valuation ring, and so you could get deeper chains of primes. Feel free to drop me an email again if you think there'll be a lot to talk about :) $\endgroup$ – rschwieb Jul 16 '15 at 18:14