A nice way to do this calculation can be extracted from the papers

[ABS] Atiyah, Bott, Shapiro: ''Clifford modules''

[AtKR] Atiyah: ''K-Theory and Reality''.

Let $A_n$ be the group of $Cl^n$-modules, modulo $Cl^{n+1}$-modules, as introduced by Atiyah, Bott, Shapiro. The sum of these is a graded ring, and the main result of [ABS] is that the homomorphism $$abs:A_{\ast} \to KO^{-\ast}(\ast)$$ they construct is a ring isomorphism. That $A_{\ast} = Z [\eta, \lambda, \beta] /(2 \eta, \eta^3, \lambda \eta, \lambda^4- 2 \beta)$ is done using linear algebra in [ABS]. The original proof in [ABS] used the known structure of $KO^{-\ast}$. In [AtKR], a simple proof of real periodicity is given, and using ideas from [AtKR], one can give a proof that $abs$ is an iso without using that knowledge, and thereby compute the $KO$-groups.

The ingredients one needs are

1. $8$-periodicity, or more precisely, that multiplication by $abs (\beta )\in KO^{-8}$ is an isomorphism.
2. The knowledge of the complex $K$-groups.
3. That the complexification $KO^0 \to K^0$ is an isomorphism.
4. That $KO^{-1} =Z/2$, generated by $abs(\eta)$. This is the easy fact that $O(n)$ has two components, in disguise.
5. The long exact sequence (proven in [AtKR], § 3)

$$\ldots KO^{1-q} \stackrel{\eta}{\to} KO^{-q} \to K^{-q} \to KO^{2-q} \ldots,$$

the map to complex $K$-theory is complexification, and we write $\eta:= abs (\eta)$ and use the same letter for the multiplication by $\eta$.

First look at the piece $$ K^{-3} \to KO^{-1} \to KO^{-2} \to K^{-2} \to KO^0 \to KO^{-1} \to K^{-1} \to KO^1\to KO^0 \to K^0$$

As $KO^0 \to K^0$ is an iso and $K^{-1}=0$, you get that $KO^1 =0$. Since $KO^{-1}=Z/2$, the map $Z= K^{-2} \to KO^{0} = Z$ must be multiplication by $\pm 2$ and hence be injective. Therefore, multiplication by $\eta$ is surjective $KO^{-1} \to KO^{-2}$. Since $K^{-3}=0$, it is also injective and you get $KO^{-2} = Z/2$, the element $\eta^2 $ is nonzero. Continue with

$$ 0= K^{-5} \to KO^{-3} \to KO^{-4} \to K^{-4} \to KO^{-2}\to KO^{-3} \to K^{-3}=0$$

Hence $\eta: KO^{-2} \to KO^{-3}$ is surjective, but as the only nonzero element of $KO^{-2}$ is $\eta^2$ and since $\eta^3 =0$ (this follows from the corresponding relation in the algebraic model $A_{\ast}$), $KO^{-2} \to KO^{-3}$ is also null. Thus $KO^{-3}=0$. It follows that $KO^{-4}=Z$, and that $KO^{-4} \to K^{-4}$ is multiplication by $\pm 2$. The last two portions of the long exact sequence are

$$ 0=K^{-7} \to KO^{-5} \to KO^{-6} \to K^{-6} \to KO^{-4} \to KO^{-5} \to K^{-5}=0 $$

and

$$ KO^{-8} \stackrel{\cong}{\to} K^{-8} \to KO^{-6} \to KO^{-7} \to K^{-7}=0. $$

We have seen that $KO^{-7}=KO^1 =0$, thus $KO^{-6}=0$ by the second sequence. This shows that $KO^{-5} $ (use first sequence) and completes the argument.

A nice way to do this calculation can be extracted from the papers

[ABS] Atiyah, Bott, Shapiro: ''Clifford modules''

[AtKR] Atiyah: ''K-Theory and Reality''.

Let $A_n$ be the group of $Cl^n$-modules, modulo $Cl^{n+1}$-modules, as introduced by Atiyah, Bott, Shapiro. The sum of these is a graded ring, and the main result of [ABS] is that the homomorphism $$abs:A_{\ast} \to KO^{-\ast}(\ast)$$ they construct is a ring isomorphism. That $A_{\ast} = Z [\eta, \lambda, \beta] /(2 \eta, \eta^3, \lambda \eta, \lambda^4- 2 \beta)$ is done using linear algebra in [ABS]. The original proof in [ABS] used the known structure of $KO^{-\ast}$. In [AtKR], a simple proof of real periodicity is given, and using ideas from [AtKR], one can give a proof that $abs$ is an iso without using that knowledge, and thereby compute the $KO$-groups.

The ingredients one needs are

1. $8$-periodicity, or more precisely, that multiplication by $abs (\beta )\in KO^{-8}$ is an isomorphism.
2. The knowledge of the complex $K$-groups.
3. That the complexification $KO^0 \to K^0$ is an isomorphism.
4. That $KO^{-1} =Z/2$, generated by $abs(\eta)$. This is the easy fact that $O(n)$ has two components, in disguise.
5. The long exact sequence (proven in [AtKR], § 3)

$$\ldots KO^{1-q} \stackrel{\eta}{\to} KO^{-q} \to K^{-q} \to KO^{2-q} \ldots,$$

the map to complex $K$-theory is complexification, and we write $\eta:= abs (\eta)$ and use the same letter for the multiplication by $\eta$.

First look at the piece $$ K^{-3} \to KO^{-1} \to KO^{-2} \to K^{-2} \to KO^0 \to KO^{-1} \to K^{-1} \to KO^1\to KO^0 \to K^0$$

As $KO^0 \to K^0$ is an iso and $K^{-1}=0$, you get that $KO^1 =0$. Since $KO^{-1}=Z/2$, the map $Z= K^{-2} \to KO^{0} = Z$ must be multiplication by $\pm 2$ and hence be injective. Therefore, multiplication by $\eta$ is surjective $KO^{-1} \to KO^{-2}$. Since $K^{-3}=0$, it is also injective and you get $KO^{-2} = Z/2$, the element $\eta^2 $ is nonzero. Continue with

$$ 0= K^{-5} \to KO^{-3} \to KO^{-4} \to K^{-4} \to KO^{-2}\to KO^{-3} \to K^{-3}=0$$

Hence $\eta: KO^{-2} \to KO^{-3}$ is surjective, but as the only nonzero element of $KO^{-2}$ is $\eta^2$ and since $\eta^3 =0$ (this follows from the corresponding relation in the algebraic model $A_{\ast}$), $KO^{-2} \to KO^{-3}$ is also null. Thus $KO^{-3}=0$. It follows that $KO^{-4}=Z$, and that $KO^{-4} \to K^{-4}$ is multiplication by $\pm 2$. The last two portions of the long exact sequence are

$$ 0=K^{-7} \to KO^{-5} \to KO^{-6} \to K^{-6} \to KO^{-4} \to KO^{-5} \to K^{-5}=0 $$

and

$$ KO^{-8} \stackrel{\cong}{\to} K^{-8} \to KO^{-6} \to KO^{-7} \to K^{-7}=0. $$

We have seen that $KO^{-7}=KO^1 =0$, thus $KO^{-6}=0$ by the second sequence. This shows that $KO^{-5}=0$ (use first sequence) and completes the argument.