The following classes are of a slightly different flavour because they depend on the additional choice of a connection.
Assume that $E\to B$ carries a flat connection $\nabla$. Then the Kamber-Tondeur classes are obstructions against the existence of a $\nabla$-parallel
metric on $E$. In the case of a complex line bundle, the first Kamber-Tondeur class is the only obstruction.
The Cheeger-Simons differential characters of a vector bundle $E\to B$ with connection $\nabla$ are obstructions against a parallel trivialisation. For a complex line bundle, the first Cheeger-Simons class is the only obstruction (in fact, this class classifies complex line bundles with connections).
Note that the Kamber-Tondeur classes can be interpreted as the imaginary parts of the Cheeger-Simons differential characters.
I will explain that as long as $n>2$ the real vector bundles $\Omega^{even}$ and $\Omega^{odd}$ over $M$ are isomorphic.
If $n>2$ then $dim(\Omega^{even}) = dim(\Omega^{odd}) = 2^{n-1} > n$ and so $\Omega^{even}$ and $\Omega^{odd}$ are isomorphic if and only if they are stably isomorphic. This follows from obstruction theory, applied to the map $BO(2^{n-1}) \to BO$. We are therefore left with analysing the real $K$-theory class
$$\Omega^{even} - \Omega^{odd} \in KO^0(M).$$
As Meier's comment points out, the complex version of this would be the complex $K$-theory Euler class of $TM$, and we can take inspiration from that construction as follows.
For a real vector space $V$ consider the chain complex of vector bundles
$$V \times \Lambda^0 V \overset{(v,w) \mapsto (v, v \wedge w)}\longrightarrow V \times \Lambda^1 V \overset{(v,w) \mapsto (v, v \wedge w)}\longrightarrow V \times \Lambda^2 V \longrightarrow \cdots,$$
which is exact over every point other than $0 \in V$. This defines (see e.g. Definition 9.23 of Spin Geometry) a compactly-supported real $K$-theory class
$$v_V \in KO^0_c(V).$$
This construction can be done fibrewise to any vector bundle, so in particular gives a class
$$v_{TM} \in KO^0_c(TM),$$
which we can consider as a class in $\widetilde{KO}^0(Th(TM))$, the reduced real $K$-theory of the Thom space of $TM$. Pulling $v_{TM}$ back along the zero-section $s_0 : M \to Th(TM)$ gives the complex of vector bundles
$$\Omega^0 \overset{0}\longrightarrow \Omega^1 \overset{0}\longrightarrow \Omega^2 \overset{0}\longrightarrow \cdots$$
over $M$, which represents the class $\Omega^{even} - \Omega^{odd} \in KO^0(M)$ that we are interested in.
On the other hand, the inclusion of a tangent fibre $S^n \to Th(TM)$ is $n$-connected, so there is a factorisation up to homotopy
$$s_0 : M \overset{q}\longrightarrow S^n \longrightarrow Th(TM)$$
(the map $q$ has degree $\chi(M)$, but that will not matter for the argument). We obtain the equation
$$\Omega^{even} - \Omega^{odd} = q^*(v_{\mathbb{R}^n}) \in KO^0(M).\tag{1}\label{eq}$$
We now turn to determining the class $v_{\mathbb{R}^n} \in KO^0_c(\mathbb{R}^n) = \widetilde{KO}^0(S^n) = KO^{-n}$. It is easy to see from the construction in terms of exterior powers that
$$v_{\mathbb{R}^n} = (v_{\mathbb{R}^1})^n \in KO^{-n},$$
and not hard to see that $v_{\mathbb{R}^1} = \eta \in KO^{-1} = \mathbb{Z}/2\{\eta\}$. It follows that $v_{\mathbb{R}^n}=0$ for $n > 2$, as $\eta^3 \in KO^{-3} = 0$, and combining this with \eqref{eq} proves the claimed result.
Addendum. It is interesting to see what happens for $n=2$. As explained in the question, over an orientable surface the Euler class distinguishes $\Omega^{even}$ and $\Omega^{odd}$, but one can still wonder about them as stable vector bundles. The analysis above, including the fact that $q$ has mod 2 degree $\chi(M)$, shows that
$$\Omega^{even} - \Omega^{odd} = \chi(M) \cdot p^*(\eta^2) \in KO^0(M),$$
where $p : M \to S^2$ is the map that collapses the complement of a ball. As $\eta^2$ has order 2, this shows that these bundles are stably isomorphic if $\chi(M)$ is even.
On the other hand taking $M = \mathbb{RP}^2$ we have $\Omega^{even} = \mathbb{R} \oplus L$, where $L$ is the tautological line bundle, and $\Omega^{odd} = T\mathbb{RP}^2 \cong_{stable} L^{\oplus 3} - \mathbb{R}$. These are not stably isomorphic, as they have different second Stiefel--Whitney classes: this implies that $\chi(\mathbb{RP}^2) \cdot p^*(\eta^2) = p^*(\eta^2) \neq 0 \in KO^0(\mathbb{RP}^2)$, which is indeed the case.
Best Answer
Here's the argument I know that avoids spectral sequences, based on the little-known space $G/N(T)$.
In between $T$ and $G$ is $N(T)$. Note that $EG$ "is an" $ET$ and $EN(T)$, since it's contractible and $T,N(T)$ act freely on it, so we can identify $BT, BN(T)$ with $EG/T, EG/N(T)$.
Now consider the two maps $EG/T \to EG/N(T) \to EG/G$, with fibers $W = N(T)/T$ and $G/N(T)$ respectively. The first case divides by a free action of $W$, so we can identify $H^\ast(BN(T);{\mathbb Q}) = H^\ast(BT; {\mathbb Q})^W$ by pushing and pulling. (Actually we only need to invert $|W|$, and generally less; for $G=U(n)$ it's true over $\mathbb Z$.) In particular, there is only even cohomology.
So let's look at the space $G/N(T) = (G/T)/W$. The space $G/T$ has a Bruhat decomposition, hence only even-degree cohomology (even over $\mathbb Z$), which you can prove via Morse theory on a generic adjoint orbit if you don't want to bring in algebraic geometry, and its Euler characteristic is $|W|$.
Hence the space $(G/T)/W$ has (rationally) only even-degree cohomology, and Euler characteristic $1$. So it has the rational cohomology of a point! For $G=SU(2)$ this space is ${\mathbb RP}^2$.
By a particularly trivial application of Leray-Hirsch (which I think is the only remainder of the spectral sequence argument Mark Grant gave), $H^\ast(EG/G; {\mathbb Q}) \cong H^\ast(EG/N(T); {\mathbb Q})$.
(Oops: I guess this answer isn't so different from Ralph's.)