The well known (Mal'tsev) conditions that characterize certain properties of the congruence lattice of an algebra. The existence of a 3-ary term $p$ together with familiar identities $p(x,y,y) \sim x \sim p(y,y,x)$, is equivalent to congruence permutability, for example. And imilar conditions exist for other lattice theoretic properties, such as distributivity and modularity.
Is there a term-condition, or an identity, that characterizes (or implies) varieties where every algebra has a self-dual congruence lattice?
A second question is if there is a Mal'tsev condition that implies that the every algebra has a complemented congruence lattice.
For example, the congruence lattice of a finite abelian group is self-dual. Or the congruence lattice of a finite dimensional vector space.
This question is about the following two properties of varieties:
The property that all members of $\mathcal V$ have self-dual congruence lattices.
The property that all members of $\mathcal V$ have complemented congruence lattices.
As Pedro observed, if all members have complemented congruence lattices, then the subdirectly irreducible members must be simple [in a nonsimple SI the monolith has no complement]. We can also say something about the SI's under the assumption that all members have self-dual congruence lattices.
First note that if all members have self-dual congruence lattices, then any interval in a congruence lattice is also self-dual. This follows from repeated use of self-duality together with the fact that upper intervals are congruence lattices of quotients.
Now assume that $\mathcal V$ has the property that its members have self-dual congruence lattices. The bottom element of the congruence lattice of an SI ${\mathbf S}$ is completely meet irreducible, so the top element must be completely join irreducible. By the observation of the last paragraph, every nontrivial upper interval is self dual. Since these intervals $[\sigma,1]$ have completely join irreducible top element, they must also have completely meet irreducible bottom element. Consequently every non-top element in the congruence lattice of ${\mathbf S}$ is completely meet irreducible, and similarly every non-bottom element is completely join irreducible. The only algebraic lattices where every non-top element is completely meet irreducible and every non-bottom element is completely join irreducible are finite chains.
Conclusion: If every algebra in ${\mathcal V}$ has a self-dual congruence lattice, then the congruence lattice of any SI in $\mathcal V$ is a finite chain.
Continuing to assume that members of $\mathcal V$ have self-dual congruence lattices, if ${\mathbf A}\in {\mathcal V}$ is nontrivial and $\alpha$ is a proper congruence on $\mathbf A$, then $\alpha$ can be enlarged to $\beta$ so that ${\mathbf A}/\beta$ is SI. By the result of the last paragraph, we may enlarge $\beta$ further to $\gamma$ if necessary where ${\mathbf A}/\gamma$ is simple. This shows that every proper congruence $\alpha$ on a nontrivial algebra in $\mathcal V$ can be enlarged to a coatom $\gamma$. By self-duality we get that every nonzero congruence must lie above an atom. This is a strong property to impose throughout a variety. It is investigated in this paper:
Atomicity and nilpotence, Canad. J. Math. 42(1990), 365-382.
A result found there is that if every nonzero congruence on every nontrivial algebra in $\mathcal V$ lies above an atom, then the ascending central series for any ${\mathbf A}\in {\mathcal V}$ reaches the top congruence, although perhaps after transfinitely many steps.
What it means here is that if we are in the case where $\mathcal V$ has self-dual congruence lattices, then SI's in $\mathcal V$ have congruence lattices that are finite chains, and the SI algebras themselves are nilpotent. (It is plausible that the SI's must even be abelian, but I see how to prove this only for congruence modular varieties at the moment.) If we are in the case where $\mathcal V$ has complemented congruence lattices, then SI's in $\mathcal V$ are simple and, as one can show by following the argument in the above paper, the SI's are abelian. In this case $\mathcal V$ is an abelian variety.
Now for the answers these questions: Is there is a Maltsev condition that implies that the every algebra has a complemented congruence lattice? and Is there is a Maltsev condition that implies that the every algebra has a self-dual congruence lattice?
Claim: The only such Maltsev conditions are those expressing that the variety is trivial.
This Claim rests on the following Observation: If $\Sigma$ is a Maltsev condition that is satisfied by a nontrivial variety, then it is satisfied by a nontrivial discriminator variety. [If $\mathcal V$ satisfies $\Sigma$, just add the discriminator $d$ to each member of $\mathcal V$ and generate a variety ${\mathcal V}^d$ with the resulting algebras. ${\mathcal V}^d$ is a discriminator variety satisfying any Maltsev condition true in ${\mathcal V}$.]
To see how the Observation implies the Claim, assume that $\Sigma$ is a Maltsev condition implying either that congruence lattices are self-dual or implying that congruence lattices are complemented. Let $\mathcal V$ satisfy $\Sigma$. The discriminator variety ${\mathcal V}^d$ satisfies $\Sigma$, so its SI's in ${\mathcal V}^d$ are nilpotent. But the only nilpotent members of a discriminator variety are trivial, so ${\mathcal V}^d$ is trivial. $\mathcal V$ must also be trivial.