Let $\Sigma \to S^{2n}$ be the spinor bundle, then one has \begin{align*} e(TS^{2n}) &= \frac{(-1)^{n-1}}{(n-1)!} \left( c_n(\Sigma^+)-c_n(\Sigma^-) \right),\\ 0 &= c_n(\Sigma). \end{align*}

Proof: The Atiyah-Singer index theorem says \begin{align*} \chi(M) &= \int_M \hat{A}(TM)ch(\Sigma^+ - \Sigma^-)=\int_M e(TM) \\ sign(M) &= \int_M \hat{A}(TM)ch(\Sigma) = \int_M L(TM) \end{align*}

Since $M=S^{2n}$ has stably trivial tangent bundle one has $\hat{A}(TM) = 1 = L(TM)$. The only thing contributing to the first integral is some rational multiple of $c_n(\Sigma^+) - c_n(\Sigma^-)$.

Now I claim for any complex vector bundle $E \to S^{2n}$ one has $$ch(E) = \frac{(-1)^{n-1}}{(n-1)!} c_n(E).$$

To see this, let $x_1,...,x_n$ denote chern roots for $E$. We have by definition $$ch(E)[S^{2n}]= \left( \sum\limits_{i=1}^{n} e^{x_i}\right) [S^{2n}] = \frac{1}{n!}\left(\sum\limits_{i=1}^n x_i^n\right)[S^{2n}].$$

Now we have to search for the coefficient of $c_n$ in the (unique) polynomial $P \in \mathbb{Z}[c_1,...,c_n]$ with $$P(c_1,\ldots,c_n)= x_1^n + \ldots + x_n^n$$ (where $c_i$ is the $i$-th elemtary symmetric polynomial in $n$ variables). Specializing to negatives of roots of unity namely $$T^n - 1 = (T + x_1)...(T+x_n) = T^n + c_{1} T^{n-1} + ... + c_{n-1} T +c_n$$ we can see that this coefficient is $- (-1)^n n$.

Let $\Sigma \to S^{2n}$ be the spinor bundle, then one has \begin{align*} e(TS^{2n}) &= \frac{(-1)^{n}}{(n-1)!} \left( c_n(\Sigma^+)-c_n(\Sigma^-) \right),\\ 0 &= c_n(\Sigma). \end{align*}

Proof: The Atiyah-Singer index theorem for spin manifolds $M^{2n}$ says \begin{align*} \chi(M) &= (-1)^{n}\int_M \hat{A}(TM)ch(\Sigma^+ - \Sigma^-)=\int_M e(TM) \\ sign(M) &= \int_M \hat{A}(TM)ch(\Sigma) = \int_M L(TM) \end{align*} Since $M=S^{2n}$ has stably trivial tangent bundle one has $\hat{A}(TM) = 1 = L(TM)$. The only thing contributing to the first integral is some rational multiple of $c_n(\Sigma^+) - c_n(\Sigma^-)$.

Now I claim for any complex vector bundle $E \to S^{2n}$ one has $$ch(E) = \frac{(-1)^{n-1}}{(n-1)!} c_n(E).$$

To see this, let $x_1,...,x_n$ denote chern roots for $E$. We have by definition $$ch(E)[S^{2n}]= \left( \sum\limits_{i=1}^{n} e^{x_i}\right) [S^{2n}] = \frac{1}{n!}\left(\sum\limits_{i=1}^n x_i^n\right)[S^{2n}].$$

Now we have to search for the coefficient of $c_n$ in the (unique) polynomial $P \in \mathbb{Z}[c_1,...,c_n]$ with $$P(c_1,\ldots,c_n)= x_1^n + \ldots + x_n^n$$ (where $c_i$ is the $i$-th elemtary symmetric polynomial in $n$ variables). Specializing to negatives of roots of unity namely $$T^n - 1 = (T + x_1)...(T+x_n) = T^n + c_{1} T^{n-1} + ... + c_{n-1} T +c_n$$ we can see that this coefficient is $- (-1)^n n$.

EDIT: Alternative Without the Index theorem one can show (using the splitting principle) that for any real spin vector bundle $E$ of even rank $2r$ $$ ch(\Sigma^+(E) - \Sigma^-(E)) = \prod\limits_{i=1}^{r} (e^{-x_i/2} - e^{x_i/2}) $$ where $x_i$ are the chern roots of $E \otimes \mathbb{C}$. The euler class $e(E) = x_1 ... x_n$ divides this expression and one has (by definition) $$ch(\Sigma^+(E)- \Sigma^-(E)) = (-1)^r e(E) \hat{A}(E)^{-1}$$ If $E$ is stabily trivial one has $\hat{A}(E) = 1$ and we are again left with determining $ch(\Sigma^+(E) - \Sigma^-(E))$ in degree $2r$. Analogously $$ ch(\Sigma(E)^+ + \Sigma^-(E)) = \prod(e^{-x_i/2} + e^{x_i/2}) = \prod (2 + \frac{1}{4}x_i^2 + ... )$$ can be expressed only in Pontrjagin classes of $E$, so in the stably trivial case every chern class of the spinor bundle has to vanish.

As a reference: Chapter III, § 11 & 12 in Spin Geometry by Lawson and Michelsohn.