Is it true that in a (non-local) Gorenstein ring, every maximal ideal has the same height? It seems a little strange, but I don't see any reason why it shoudn't.

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No, it is not true.

Consider a discrete valuation ring $R$ with uniformizing parameter $\pi$ and residue field $k$.
Take for your Gorenstein ring the polynomial ring $A=R[X]$: since $R$ is regular, so is $A$ and thus $A$ is a fortiori Gorenstein.
Now consider the ideals $\mathfrak p=(\pi, X)$ and $\mathfrak q=(\pi X-1)$.
We have $A/\mathfrak p=k$, a field, and $A/\mathfrak q=Frac(R)$, a field too. So both $\mathfrak p$ and $\mathfrak q$ are maximal ideals. However their heights are different: $ht(\mathfrak p)=2$ (use $dim A=2$) and $ht(\mathfrak q)=1$ (use Krull's principal ideal theorem)

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I think the answer is no. Take the product of two local Gorenstein rings of different dimensions; or take any Gorenstein ring $A$ and two prime ideals $P, Q$ of different heights, and no inclusion relation between them and localize at $A\setminus (P\cap Q)$ (you get a semi-local ring whose maximal ideals are given by $P$ and $Q$).

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Very nice! [Je ne savais pas ce qu'est un anneau de Gorenstein et avais commencé à taper avant de voir votre réponse, mais motus ... J'étais parti du principe qu'un anneau régulier a de toute façon toutes les propriétés de régularité imaginables: Gorenstein, Frankenstein, ... ] –  Georges Elencwajg Apr 18 '12 at 15:52
@GeorgesElencwajg: ton exemple est très concret:). Regular $\Longrightarrow$ local complete intersection $\Longrightarrow$ Gorentstein $\Longrightarrow$ Cohen-Macaulay. –  user18119 Apr 18 '12 at 16:24
Yes, I had checked the definition in Matsumura before posting my answer. I suppose it will take some time before I have a gut feeling for Gorenstein, which seems a very weak concept. Any hint for helping me understand it? [Ah, et je vais donc avoir le plaisir de te tutoyer!] –  Georges Elencwajg Apr 18 '12 at 16:36
@GeorgesElencwajg: I'm not acquainted with Gorenstein, Frankenstein... sorry. For me, it means the dualizing complex is an invertible sheaf. [Je pense que tu me tutoyais déjà il y a longtemps] –  user18119 Apr 18 '12 at 17:01