I know how to construct all ordinals less than $\omega^3$.

Let $k$ be an infinite field.

Lemma 1: For any Noetherian ring $R$, $$\alpha( R \times k ) \geq \alpha(R)+1.$$

Proof: $R \times k$ can move to $R$ and it can also move to every element $R$ can move to, so $\alpha(R \times k) \geq \alpha(R)$.

Lemma 2: For any Noetherian ring $R$, $$\alpha\left(R \times k[x]\right) \geq \alpha(R) + \omega.$$

Proof: By inductively applying Lemma 1, every ordinal $< \alpha(R) + \omega$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k$. Hence every ordinal $< \alpha(R) + \omega$ is equal to a quotient of such a ring by a single element, which is necessarily a quotient of $R$ by zero or one elements times a finite product of copies of $k$, which is manifestly a quotient of $R \times k[x] $ by one element.

Lemma 3: For any Noetherian ring $R$, $$\alpha\left( R \times k[x,y] \times k[x]\right) \geq \alpha(R) + \omega^2$$

Proof: By inductively applying Lemma 2, every ordinal $< \alpha(R) + \omega^2$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k[x]$. So it is equal to the ordinal of a quotient of $R$ by $0$ or $1$ elements, times a finite product of copies of $k[x]$, times a finite product of quotients of $k[x]$ by one element. The first is a quotient of $R$ by $0$ or $1$ elements, the second is a quotient of $k[x,y]$ by one element, and the third, crucially, is a quotient of $k[x]$ by one element, because we can fit any finite number of zero-dimensional schemes that individually embed into the line together in the same line so that they do not intersect. Hence every ordinal less than $\alpha(R) + \omega^2$ is the ordinal of a quotient of $R \times k[x,y] \times k[x]$ by one element.

This last step prevents us from extending the argument one dimension higher, because there is no Noetherian affine scheme such that a finite union of hypersurfaces in $\mathbb A^2$ is itself a hypersurface in that scheme - at least I don't think there is.

However, by iterating Lemma 3, we can construct rings with nimbers at least $ n \omega^2$ for all natural numbers $n$, and hence verify the existence of rings with any given nimber $< \omega^3$.

Note that we have done this inductive argument without at any point upper bounding the nimber of any ring. I am guessing that at some point upper bounds for certain rings will be needed to lower bound the nimbers of other rings.

I know how to construct all ordinals less than $\omega^3$.

Let $k$ be an infinite field.

Lemma 1: For any Noetherian ring $R$, $$\alpha( R \times k ) \geq \alpha(R)+1.$$

Proof: $R \times k$ can move to $R$ and it can also move to every element $R$ can move to, so $\alpha(R \times k) \geq \alpha(R)$.

Lemma 2: For any Noetherian ring $R$, $$\alpha\left(R \times k[x]\right) \geq \alpha(R) + \omega.$$

Proof: By inductively applying Lemma 1, every ordinal $< \alpha(R) + \omega$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k$. Hence every ordinal $< \alpha(R) + \omega$ is equal to the ordinal of a quotient of such a ring by a single element, which is necessarily a quotient of $R$ by zero or one elements times a finite product of copies of $k$, which is manifestly a quotient of $R \times k[x] $ by one element.

Lemma 3: For any Noetherian ring $R$, $$\alpha\left( R \times k[x,y] \times k[x]\right) \geq \alpha(R) + \omega^2$$

Proof: By inductively applying Lemma 2, every ordinal $< \alpha(R) + \omega^2$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k[x]$. So it is equal to the ordinal of a quotient of $R$ by $0$ or $1$ elements, times a finite product of copies of $k[x]$, times a finite product of quotients of $k[x]$ by one element. The first is a quotient of $R$ by $0$ or $1$ elements, the second is a quotient of $k[x,y]$ by one element, and the third, crucially, is a quotient of $k[x]$ by one element, because we can fit any finite number of zero-dimensional schemes that individually embed into the line together in the same line so that they do not intersect. Hence every ordinal less than $\alpha(R) + \omega^2$ is the ordinal of a quotient of $R \times k[x,y] \times k[x]$ by one element.

This last step prevents us from extending the argument one dimension higher, because there is no Noetherian affine scheme such that a finite union of hypersurfaces in $\mathbb A^2$ is itself a hypersurface in that scheme - at least I don't think there is.

However, by iterating Lemma 3, we can construct rings with nimbers at least $ n \omega^2$ for all natural numbers $n$, and hence verify the existence of rings with any given nimber $< \omega^3$.

Note that we have done this inductive argument without at any point upper bounding the nimber of any ring. I am guessing that at some point upper bounds for certain rings will be needed to lower bound the nimbers of other rings.