Tetsu Nishimoto kindly performed the computation, and allowed me to reproduce it here. --Yuji

Proposition:
The mod-2 cohomology $H^*(BSs(16m);\mathbb Z/2)$ of the classifying space of the Lie group $Ss(16m)$, is isomorphic to the following algebra up degree $ \leq 11$:
$$
\mathbb Z/2[x_2, x_3, x_5, x_9, y_4, y_6, y_7, y_{10}, y_{11}]
/(x_2y_7+x_3y_6+x_5y_4+x_2x_3y_4).
$$
Within $\ast \leq 11$, the action of $Sq^k$ is given by
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
& Sq^1 & Sq^2 & Sq^3 & Sq^4 & Sq^5 \\
\hline
x_2 & x_3 & x_2^2 & & & \\
x_3 & 0 & x_5 & x_3^2 & & \\
x_5 & x_3^2 & 0 & 0 & x_9 & x_5^2 \\
x_9 & x_5^2 & 0 & & & \\
y_4 & 0 & y_6 & y_7 & y_4^2 & \\
y_6 & y_7 & 0 & 0 & y_{10} & y_{11} \\
y_7 & 0 & 0 & 0 & y_{11} & \\
y_{10} & y_{11} & & & & \\
\hline
\end{array}.
$$

Let us describe the outline of the computation.
First we need to quote the structure of $H^*(Ss(16m);\mathbb Z/2)$ as a Hopf algebra from
Proposition 4.1 of

Hopf Algebra Structure of mod 2 Cohomology of Simple Lie groups

K. Ishitoya, A. Kono and H. Toda

Publ. RIMS, Kyoto Univ. *12* (1976) 141-167 electric version

The proposition states that, in the range $\ast \leq 10$,
$H^*(Ss(16m);\mathbb Z/2)$ is isomorphic as an algebra to
$$
\Delta (w_3, w_5, w_6, w_7, w_9, w_{10}) \otimes \mathbb Z/2[\bar{v}]
$$
where $\deg w_i = i$, $\deg \bar{v} = 1$.
The generators other than $w_7$ are primitive, while the coproduct of $w_7$ is given by
$$
\bar\psi (w_7) = \bar{v} \otimes w_6 + \bar{v}^2 \otimes w_5
+ \bar{v}^4 \otimes w_3.
$$
The action of $Sq^k$ within the range $\ast \leq 10$ is given by
$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
& Sq^1 & Sq^2 & Sq^3 & Sq^4 & Sq^5 \\
\hline
w_3 & 0 & w_5 & w_6 = w_3^2 & & \\
w_5 & w_6 & 0 & 0 & w_9 & w_{10} = w_5^2 \\
w_6 & 0 & 0 & 0 & w_{10} & \\
w_7 & m\bar{v}^8 & w_9 & w_{10} & & \\
w_9 & w_{10} & & & & \\
\hline
\end{array}.
$$

Next, we consider the Rothenberg-Steenrod spectral sequence
$$
E_2 = \mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)}
(\mathbb Z/2, \mathbb Z/2)
\Longrightarrow H^*(BSs(16m);\mathbb Z/2).
$$

We first need to compute the $E_2$ term.
Here we use May's spectral sequence
$$
E'_1=\mathrm{Cotor}_{A'}(k,k) \Longrightarrow \mathrm{Cotor}_{A}(k,k).
$$ Here, $A'$ is a Hopf algebra such that it is isomorphic as an algebra with $A'$ such that every generator is primitive.
When the characteristic of $k$ is $2$, $\mathrm{Cotor}_{A'}(k,k)$ is a polynomial ring whose generators are in one-to-one correspondence with the primitive elements of $A'$.
Here we take $k=\mathbb{Z}/2$ and $A=H^*(Ss(16m);\mathbb Z/2)$.
Then, up to degree 11, we have
$$
\mathrm{Cotor}_{A'}(k,k) = \mathbb{Z}/2[[v],[v^2],[v^4],[v^8],[w_3],[w_5],[w_6],[w_7],[w_9],[w_{10}]]
$$
The differential at $E_2$ is given by
$$
d_2([w_7]) = [v][w_6]+[v^2][w_5].
$$
Note that from the construction of May's spectral sequence the term $[v^4][w_3]$ vanish.
As all the other differentials are zero, $E'_\infty$ up to degree 11 is isomorphic to
$$
\mathbb Z/2[[v],[v^2],[v^4],[v^8],[w_3],[w_5],[w_6],[w_9],[w_{10}]]/([v][w_6]+[v^2][w_5]).
$$
It is easily seen that in $\mathrm{Cotor}_A(k,k)$ the relation corresponding to $[v][w_6]+[v^2][w_5]$ is $[v][w_6]+[v^2][w_5]+[v^4][w_3]$. Therefore $\mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)}
(\mathbb Z/2, \mathbb Z/2) $ is given by
$$
\mathbb Z/2 [[\bar{v}], [\bar{v}^2], [\bar{v}^4], [\bar{v}^8],
[w_3], [w_5], [w_6], [w_9], [w_{10}]]/
([\bar{v}][w_6]+[\bar{v}^2][w_5]+[\bar{v}^4][w_3])
$$ up to degree 11 as algebras.

We note here that $[\bar{v}^{2^j}] \in E_2^{1,2^j}$, $[w_i] \in E_2^{1,i}$,
and that these generators all correspond to primitive elements.
When the degrees are higher this is not necessarily the case. The relation came from the coproduct of $w_7$, as we saw above.

The differentials are given in the range $r \geq 2$ as
\begin{align*}
& d_r : E_r^{1,1} \longrightarrow E_r^{1+r,1-(r-1)} =
E_2^{1+r,1-(r-1)} = 0, \\
& d_r : E_r^{1,3} \longrightarrow E_r^{1+r,3-(r-1)} =
E_2^{1+r,3-(r-1)} = 0.
\end{align*}
Therefore, $[\bar{v}]$ and $[w_3]$ are permanent cycles.
In general, when $x$ is a permanent cycle, for any cohomology operation $\theta$
$\theta x$ is also a permanent cycle.
Other generators can be written as
\begin{align*}
& [\bar{v}^2] = Sq^1[\bar{v}],
\quad
[\bar{v}^4] = Sq^2Sq^1[\bar{v}],
\quad
[\bar{v}^8] = Sq^4Sq^2Sq^1[\bar{v}], \\
& [w_5] = Sq^2[w_3],
\quad
[w_6] = Sq^3[w_3],
\quad
[w_9] = Sq^4Sq^2[w_3],
\quad
[w_{10}] = Sq^5Sq^2[w_3]
\end{align*}
and therefore these are also permanent cycles.
Therefore, up to degree 11, we have $E_{\infty} = E_2$.

Let us now define elements of $H^*(BSs(16m);\mathbb Z/2)$.
Let $x_2$ be a representative of $[\bar{v}]$,
and $y_4$ be a representative of $[w_3]$.
$x_2$ is uniquely determined but there are two elements $y_4$ and $y_4+x_2^2$ representing $[w_3]$. This freedom is used below when we fix the relations.
Let us further set
\begin{align*}
& x_3 = Sq^1 x_2,
\quad
x_5 = Sq^2 x_3,
\quad
x_9 = Sq^4 x_5, \\
& y_6 = Sq^2 y_4,
\quad
y_7 = Sq^1 y_6,
\quad
y_{10} = Sq^4 y_6,
\quad
y_{11} = Sq^1 y_{10}
\end{align*}
then they are representatives of $[\bar{v}^2]$, $[\bar{v}^4]$, $[\bar{v}^8]$,
$[w_5]$, $[w_6]$, $[w_9]$, $[w_{10}]$, respectively.
Using the Adem relation, we can determine how $Sq^k$ acts on these elements, giving the table shown above.

Finally let us determine the relation in $H^*(BSs(16m),\mathbb Z/2)$ corresponding to
the relation $[\bar{v}][w_6]+[\bar{v}^2][w_5]+[\bar{v}^4][w_3]$ of
$\mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)}
(\mathbb Z/2, \mathbb Z/2)$.
When $k \geq 3$, the basis of $E_2^{k,9-k}$ can be given by
$$
x_2^2x_5, \quad x_3^3, \quad x_2^3x_3, \quad x_2x_3y_4
$$
and therefore the degree-9 relation $r$ in $H^*(BSs(16m);\mathbb Z/2)$ can be given by
$$
r = x_2y_7 + x_3y_6 + x_5y_4 + a_1x_2^2x_5 + a_2x_3^3
+ a_3x_2^3x_3 + a_4x_2x_3y_4
\quad (a_i \in \mathbb Z/2).
$$
From
$$
Sq^1 r = (1+a_4)x_3^2y_4 + (a_1+a_3)x_2^2x_3^2
$$
we have $a_4 = 1$ and $a_1 = a_3$. From
\begin{align*}
Sq^2 r & = x_2^2y_7 + (a_1+a_2)x_3^2x_5 + a_1x_2^4x_3
+ a_1x_2x_3^3 + a_1x_2^3x_5 + x_2^2x_3y_4 + x_2x_5y_4
+ x_2x_3y_6 \\
& = (a_1+a_2)x_3^2x_5
+ (a_1+a_2)x_2x_3^3
\end{align*}
we have $a_1 = a_2$. Then the relation is given by
$$
r = x_2y_7 + x_3(y_6+a_1x_3^2) + (x_5+x_2x_3)(y_4+a_1x_2^2).
$$
When $a_1 = 1$, we exchange $y_4$ by $y_4+x_2^2$.
Then $y_6$ is exchanged with $y_6+x_3^2$ and $y_{10}$ is exchanged with $y_{10}+x_5^2$, while all the other generators are fixed.
The relation then becomes
$$
r = x_2y_7 + x_3y_6 + (x_5+x_2x_3)y_4
$$
and has the same form as the relation for the case $a_1 = 0$.
This completes the determination of the relation.