It seems that one can obtain the additive structure of rational cohomology
without too much effort (in no way have I checked this carefully so caveat
lector applies). As Allen noticed, for rational cohomology it is enough to
compute $T$-equivariant cohomology and then take $\Sigma_m$-invariants (if this is to
work also for integral cohomology a more careful analysis would have to be made I think). Now, the space $Tor^m_C$ of length $m$ quotients of $\mathcal O^m_C$ (I use $C$
as everything I say will work for any smooth and proper curve $C$) is smooth and proper so we may use the Bialinski-Birula analysis (choosing a general $\rho\colon \mathrm{G}_m \to T$) and we first look at the fixed point locus of $T$.

Now, for every sequence $(k_1,\ldots,k_m)$ of non-negative integers with
$k_1+\ldots+k_m=m$ gives a map $S^{k_1}C\times\cdots\times S^{k_m}C \to Tor^m_C$, where $S^kC$ is the
symmetric product interpreted as a Hilber scheme, and the map takes $(\mathcal
I_1,\ldots,\mathcal I_m)$ to $\bigoplus_i\mathcal I_i \hookrightarrow \mathcal O^m_C$. It is
clear that this lands in the $T$-fixed locus and almost equally clear that this
is the whole $T$-fixed locus (any $T$-invariant submodule must be the direct sum of its
weight-spaces).

We can now use $\rho$ to get a stratification parametrised by the sequences
$(k_1,\ldots,k_m)$. Concretely, the tangent space of $Tor^m_C$ to a point of
$S^{k_1}C\times\cdots\times S^{k_m}C$ has character $(k_1\alpha_1+\cdots+k_m\alpha_m)\beta$, where
$\beta=\sum_i\alpha_i^{-1}$ (and we think of characters as elements of the group ring
of the character group of $T$ and the $\alpha_i$ are the natural basis elements of
the character group). This shows that $\rho$ can for instance be chosen to be $t \mapsto
(1,t,t^2,\ldots,t^{m-1})$. In any case the stratum corresponding to $(k_1,\ldots,k_m)$ has
as character for its tangent space the characters on the tangent space on which
$T'$ is non-negative and its normal bundle consists of those on which $T'$ is
non-negative (hence with the above choice, the character on the tangent space is
$\sum_{i\geq j}k_i\alpha_i\alpha_j^{-1}$ and in particular the dimension of the stratum is
$\sum_iik_i$). We now have that each stratum is a vector bundle of the
corresponding fixed point locus so in particular the equivariant cohomology of
it is the equivariant cohomology of the fixed point locus and in particular is
free over the cohomology ring of $T$. Furthermore, if we build up the cohomology
using the stratification (and the Gysin isomorphism), at each stage a long exact
sequence splits once we have shown that the top Chern class of the normal bundle
is a non-zero divisor in the cohomology of the equivariant cohomology of the
component of the fixed point locus (using the Atiyah-Bott criterion).

However, the non-zero divisor condition seems more or less automatic (and that
fact should be well-known): The normal bundle $\mathcal N$ of the
$(k_1,\ldots,k_m)$-part, $F$ say, of the fixed point locus splits up as direct sum
$\bigoplus_\alpha \mathcal N_\alpha$, where $T$ acts by the character $\alpha$ on $\mathcal N_\alpha$. Then
the equivariant total Chern class of $\mathcal N_\alpha$ inside of
$H^\ast_T(F)=H^\ast(F)\bigotimes H^\ast_T(pt)$ is the Chern polynomial of $\mathcal N_\alpha$
as ordinary vector bundle evaluated at $c_1(\alpha)=\alpha\in H^2_T(pt)$. Hence, if we quasi-order
the characters of $T$ by using $\beta \mapsto -\rho(\beta)$ (``quasi'' as many characters get the
same size), then as $-\rho(\alpha)>0$ (because $\alpha$ appears in the normal bundle) we get
that $1\otimes\alpha^{n_\alpha}\in H^\ast_T(F)$, where $n_\alpha$ is the rank of $\mathcal N_\alpha$, is the term of
$c_n(\mathcal N_\alpha)$ of largest order. Hence, we get that for the top Chern class
of $\mathcal N$ which is the product $\prod_\alpha c_{n_\alpha}(\mathcal N_\alpha)$ its term of
largest order has $1$ as $H^\ast(F)$-coefficient and hence is a non-zero divisor.

As the cohomology of $S^nC$ is torsion free we get that all the involved
cohomology is also torsion free and everything works over the integers but as
I've said to go integrally from $T$-equivariant cohomology to
$\mathrm{GL}_m$-equivariant cohomology is probably non-trivial.

If one wants to get a hold on the multiplicative structure one could use that
the fact that the Atiyah-Bott criterion works implies that that $H^\ast_T(Tor^m_C)$
injects into the equivariant cohomology of the fixed point locus. The algebra
structure of the cohomology of $S^nC$ is clear (at least rationally) so we get
an embedding into something with known multiplicative structure. The tricky
thing may be to determine the image. We do get a lower bound for the image by
looking at the ring generated by the Chern classes of of the tautological bundle
but I have no idea how close that would get us to the actual image. (It is a
well-known technique anyway used in for instance equivariant Schubert calculus
so there could be known tricks.)

There is another source of elements, namely we have the map $Tor^m_C \to S^mC$. This map becomes
even better if one passes to the $\Sigma_m$-invariants.

[Later] Upon further thought I realise that the relation between $T$- and $G=GL_m$-equivariant cohomology
is simpler than I thought. The point is (and more knowledgeable people certainly know this) that the map $EG\times_TX \to EG\times_GX$ is a $G/T$ bundle and $G$ being special $H^*_T(X)$ is free as a $H^*_G(X)$-module (with $1$ being one of the basis elements). That means that $H^*_G(X) \to H^*_T(X)$
is injective but more precisely $H^*_T(X)/H^*_G(X)$ is torsion free with $\Sigma_m$-action without invariants so that $H^*_G(X)$ is the ring of $\Sigma_m$-invariants of $H^*_T(X)$.

`$h:\mathcal{O}_E^{\oplus m}\to T$`

where $T$ is torsion of length 2 and such that $H^0(h)$ is an isomorphism... Then $Tor^m = Q^0/GL_m$. – Dragos Fratila Jan 11 '13 at 16:35