I have the $N$-fold tensor product of a convex combination of a pure state, i.e. $|\psi\rangle\langle\psi|$ with $|\psi\rangle$ a unit vector in a complex Hilbert space of dimension two, and the completely mixed state $\frac{\mathrm{I}_2}{2}$, with $\mathrm{I}_2$ the identity operator in two dimensions:
$$\rho(\lambda) = \left(\lambda|\psi\rangle\langle\psi|+(1-\lambda)\frac{\mathrm{I}_2}{2}\right)^{\otimes N},\\ \mathrm{with}~\lambda \in \left(0,1\right) $$
Writing w.l.o.g. $|\psi\rangle = \begin{bmatrix} \alpha\\ \beta \end{bmatrix}$ with $\alpha,\beta \in \mathbb{C}, ~|\alpha|^2+|\beta|^2 = 1$,
$$\rho(\lambda) = \left(\begin{bmatrix} \lambda|\alpha|^2+\frac{1-\lambda}{2} & \lambda\alpha\beta^*\\ \lambda\alpha^*\beta &\lambda|\beta|^2+\frac{1-\lambda}{2} \end{bmatrix}\right)^{\otimes N}$$
For $\lambda = 0$, the dimension of all the states in the form of $\rho$ with varying $|\psi\rangle$ is $2^N$, while for $\lambda = 1$ the space of all possible $\rho$ is the totally symmetric subspace, so that the dimension is given by $N+1$. A way to formalize the dimension of some space is to calculate the (lowest possible) trace of the identity operator working on that space. For example, $N = 2, \lambda = 1$ we can write our identity operator on all states of those form as
$$ \begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & \frac{1}{2} & \frac{1}{2} & 0\\ 1 & \frac{1}{2} & \frac{1}{2} & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} $$
The trace over this operator gives $2+1 = 3$, and doing the same for $N>2$ and $\lambda = 1$ you would be always able to find some operator that acts as the identity operator on those states, where $N+1$ will be the trace over that operator. We can assign the trace of that operator to the dimension of that space (see https://en.wikipedia.org/wiki/Dimension_(vector_space)#Trace).
What can you say about the dimension for intermediate values of $\lambda$? Do you immediately lose the ability to reduce the dimension for $\lambda<1$?
