This is to expand further on something I wrote in the comments. Martin wrote:

Here are some necessary conditions: [...] Are they sufficient?

I think no since a polynomial algebra $\mathbb{R}[x_1,\ldots,x_n]$ satisfies all these conditions but is not isomorphic to the algebra of smooth functions on a manifold. A short explanation for this is that smooth functions satisfy Borel's lemma while polynomial algebras don't. Here's a more detailed explanation:

Let $m_p$ the maximal ideal corresponding to a point $p\in Spec_{\mathbb{R}}A$. Then for polynomial algebras as well as algebras of the form $C^\infty(M)$, the $m_p$-completion $\hat{A}=\varprojlim A/m_p^r$ is (after fixing local coordinates) a formal power series algebra $R[[x_1,\ldots,x_n]]$. The natural map $A\to \hat A$ may be interpreted as associating to a function its Taylor expansion at $p$. One version of the lemma of Borel says that for smooth function algebras this map is surjective. In other words: for every power series I give you (even non convergent) you can find a smooth function which has it as its Taylor series. Obviously this does not hold for polynomial algebras. So this gives you another necessary condition.

In the comments I said that polynomial algebras are finitely generated while algebras $C^\infty(M)$ are not. You asked me how to see this. I don't know if there is a simpler proof, but I would apply the same lemma of Borel: finitely generated algebras have a countable basis as vector spaces. But formal power series have no countable basis, and since the map $A\to \hat A$ is surjective also $A$ cannot have a countable basis. (Obviously if you just wanted to know that polynomial algebras are not isomorphic to smooth function algebras you didn't need this anymore).

This is to expand further on something I wrote in the comments. Martin wrote:

Here are some necessary conditions: [...] Are they sufficient?

I think no since a polynomial algebra $\mathbb{R}[x_1,\ldots,x_n]$ satisfies all these conditions but is not isomorphic to the algebra of smooth functions on a manifold. A short explanation for this is that smooth functions satisfy Borel's lemma while polynomial algebras don't. Here's a more detailed explanation:

Let $m_p$ the maximal ideal corresponding to a point $p\in Spec_{\mathbb{R}}A$. Then for polynomial algebras as well as algebras of the form $C^\infty(M)$, the $m_p$-completion $\hat{A}=\lim_{r \geq 1} A/m_p^r$ is (after fixing local coordinates) a formal power series algebra $R[[x_1,\ldots,x_n]]$. The natural map $A\to \hat A$ may be interpreted as associating to a function its Taylor expansion at $p$. One version of the lemma of Borel says that for smooth function algebras this map is surjective. In other words: for every power series I give you (even non convergent) you can find a smooth function which has it as its Taylor series. Obviously this does not hold for polynomial algebras. So this gives you another necessary condition.

In the comments I said that polynomial algebras are finitely generated while algebras $C^\infty(M)$ are not. You asked me how to see this. I don't know if there is a simpler proof, but I would apply the same lemma of Borel: finitely generated algebras have a countable basis as vector spaces. But formal power series have no countable basis, and since the map $A\to \hat A$ is surjective also $A$ cannot have a countable basis. (Obviously if you just wanted to know that polynomial algebras are not isomorphic to smooth function algebras you didn't need this anymore).