The goal of my answer is to elaborate on steps 1 and 2 of the accepted answer of Steven Landsburg. In particular, I would like to make clear how a "small enough ideal" looks like and how to produce an non-reducible row like the one in Oblomov's comment. This is essentially revisiting the proof of [1, Proposition 1.9], with some K-theoretical shortcuts. I'll also show that the Bass stable range of $\mathbf{Z}[X^{\pm1}]$ is still $3$ using the very same Vaserstein recipe.
The Bass-Milnor-Serre theorem [2, Theorem 11.33] is key to understand step 1: let $F$ be a totally imaginary number field and $m$ is the number of elements of finite order in $F^{\ast}$. For each ideal $I$ in the ring of integers $R$ of $F$, the relative special Whitehead group $SK_1(R, I)$ is a cyclic group of finite order $r$. For each prime $p$, $\text{ord}_p(r)$ is the nearest integer in the interval $\lbrack 0, \text{ord}_p(m) \rbrack$ to $$\min_{\mathfrak{p}} \left\lfloor \frac{\nu_{\mathfrak{p}}(I)}{\nu_{\mathfrak{p}}(pR)} - \frac{1}{p - 1} \right\rfloor$$ where $\lfloor x \rfloor$ denotes the greatest integer $\le x$ and $\mathfrak{p}$ ranges over the prime ideals of $R$ containing $p$.
Indepedently of the above theorem, it should be clear from the definition that $SK_1(R, I)$ surjects onto $SK_1(R, J)$ if $I \subset J$. Hence $SK_1(R, I)$ increases in size as $I$ decreases with respect to inclusion. This is reflected in the above formula and it is immediate to see that for every decreasing sequence of ideals $I$, the corresponding sequence $(SK_1(R, I))_I$ stabilizes. Moreover, we can find $I$ such that $SK_1(R, I)$ has the maximal cardinality $m$ and $I$ is then small enough in the sense that any ideal contained in $I$ yields the same maximal $SK_1$ group.
We are now (almost) in position to provide and order $A$ in some $R$ as above such that $\mathbf{Z}[X]$ surjects onto $A$ and $SK_1(A)$ is not trivial. Indeed, consider $R = \mathbf{Z}[i]$ the Gauss integers and set $A = \mathbb{Z} + 4i \mathbb{Z}$. We will establish that $SK_1(A)$ surjects onto $SK_1(R, I) \simeq \mathbf{Z}/2\mathbf{Z}$, the latter isomorphism being given by Bass-Milnor-Serre's theorem. To do so, let us prove the following lemma:
Let $S$ be a one-dimensional Noetherian domain and let $I$ be an ideal of $S$ such that $S/I \simeq \mathbf{Z}/n\mathbf{Z}$ for some $n \ge 0$. Then $SK_1(S) = SK_1(S, I)$.
Proof. In the exact sequence [2, Theorem 13.20 and Example 13.22] $$K_2(S/I) \rightarrow SK_1(S, I) \rightarrow SK_1(S) \rightarrow SK_1(S/I)$$ the last term, namely $SK_1(S/I)$, is trivial since $S/I$ is finite. In addition, the image of $K_2(S/I)$ in $SK_1(S, I)$ is also trivial. Indeed $K_2(S/I)$ is generated by the Steinberg symbol $\{-1 + I, -1 + I\}_{S/I}$ because $S/I\simeq \mathbf{Z}/n\mathbf{Z}$ [2, Exercises 13A.10 and 15C.10]. As $\{-1, -1\}_S$ is a lift of the previous symbol in $K_2(S)$ our claim follows and the proof is now complete.
By the previous lemma we have $SK_1(A) = SK_1(A, I)$. Since the inclusion $A \subset R$ induces an epimorphism from $SK_1(A, I)$ onto $SK_1(R, I)$, we deduce that $SK_1(A) \neq 1$.
It is actually possible to find a non-trivial element in $SK_1(A)$ by considering the power residue symbol $\binom{4}{5 + 8i}_4 = -1$ [2, Section 11C]. This yields an non-trivial Mennick symbol $[4, 5 + 8i]$ and eventually the non-reducible row $(4, 5 + 2X, X^2 + 16)$.
Now localizing $A$ and $R$ at $x = 1 + 4i$, the same line of reasoning applies to show that $\text{sr}(\mathbb{Z}[X^{\pm 1}]) = 3$.
[1] F. Grunwald et al., "On the groups $SL_2(\mathbf{Z}[x])$ and $SL_2(k[x,y])$", 1994.
[2] B. Magurn, "An algebraic introduction to K-theory", 2002.