The following reference should be of interest :

Brewer, Montgomery, Rutter, Heinzer, *Krull dimension of polynomial rings*.

For example, they prove, see Corollary 2 p. 30, that any semi-hereditary ring (all finitely generated ideals are projective) satisfies the dimension formula above. This generalizes Seidenberg's result since a Prüfer ring is a semi-hereditary integral domain.

It might be interesting to study the class of rings $R$ satisfying the following condition : for every prime ideal $P$ of $R$ and every $n \geq 1$, we have $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$. This condition implies $\dim R[X_1,\ldots,X_n] = \dim R +n$ for every $n$ (this can be deduced from Thm 1 of this paper), but I don't know about the converse.

The authors also discuss the class of strong $S$-rings introduced by Kaplansky (see the paper for the definition). This class contains the Noetherian rings and the Prüfer rings, and is stable by localizations and quotients. Kaplansky proved that a strong $S$-ring $R$ satisfies $\mathrm{height}(P[X]) = \textrm{height}(P)$ for every prime ideal $P$ of $R$, and thus $\textrm{dim}(R[X])=\textrm{dim}(R)+1$. But a strong $S$-ring doesn't necessarily satisfy the dimension formula for every $n$. In the other direction, the authors give an example of a ring which satisfies the height formula $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$ for every prime ideal $P$, but which is not a strong $S$-ring.

Krull dimension of polynomial rings. For example, they prove, see Corollary 2 p. 30, that any semi-hereditary ring (all finitely generated ideals are projective) satisfies the dimension formula above. This generalizes Seidenberg's result since a Prüfer ring is a semi-hereditary integral domain. – François Brunault Dec 4 '12 at 16:28