I think Nik Weaver is right that the ideal mentioned is the unique maximal ideal.
This simultaneously answers both questions (since the quotient is clearly infinite dimensional). Let $\tau$ be the trace on $M_\mathcal{U}$ defined as $\tau(x_n)=\lim_{n\rightarrow \mathcal{U}}\tau_n(x_n)$ where $\tau_n$ is the normalized trace on $M_n.$ As Nik Weaver mentioned in the comments, the ideal $\{ x\in M_\mathcal{U}:\tau(x^*x)=0 \}$ is maximal. I claim this is the only maximal ideal. First we need a lemma from linear algebra

**Claim:** Let $a\in M_N$ be positive norm 1 and set $\varepsilon=\tau_N(A)>0.$ Then there are
$k=\frac{2}{\varepsilon}$ partial isometries $v_1,...,v_k\in M_N$ such that $\sum v_i^*av_i\geq \frac{\varepsilon}{2}I.$

**Proof of Claim:** Order the eigenvalues of $a$ as $a_1\geq a_2\geq\cdots \geq a_N$ and if $a_{\frac{N\varepsilon}{2}+1}<\frac{\varepsilon}{2}$ the trace is strictly less than $\varepsilon$ hence $a_i\geq \frac{\varepsilon}{2}$ for $1\leq i\leq \frac{N\varepsilon}{2}.$ Let $v_1$ project onto the $\frac{N\varepsilon}{2}$-dimensional subspace spanned by the first $\frac{N\varepsilon}{2}$ eigenvectors (corresponding to the ordering of the eigenvalues $a_i$). Then twist this projection down the line with appropriate partial isometries to obtain the claim.

**Back to the Answer:** Let $I$ be an ideal that contains a positive, norm 1 element $x=(x_n)$ such that $\tau(x)>0.$ We will show that the ideal generated by $x$ contains the identity. By replacing $(x_n)$ with an equivalent sequence we can assume each $x_n$ is positive, norm 1 and has $\tau_n(x_n)\geq\varepsilon$ for some $\varepsilon>0.$ Now just apply the above claim coordinate wise to produce $k=\frac{2}{\varepsilon}$ partial isometries $w_1,...,w_k\in M_\mathcal{U}$ so $\sum w_ixw_i^*\geq \frac{\varepsilon}{2}I.$

Operator Algebras). Could this be the only maximal ideal? $\endgroup$normbounded, as (surely?) your traces are normalised. $\endgroup$