No need to specify the ambient ring since every commutative associative ring $I$ (possibly without unit) is an ideal in a ring (namely $I\oplus\mathbf{Z}$ with multiplicative law $(a_1,n_1)(a_2,n_2)=(a_1a_2+n_1a_2+n_2a_1,n_1n_2)$).

Now call a X-ring a ring (possibly without unit) satisfying the axioms: associative, commutative, $x+x=x^2=0$ for all $x$.

So the free X-ring $R_I$ on the generators $(x_i)_{i\in I}$ has the basis over the field on 2 elements: $(x_J)_{J\in 2^I_0}$, where $2^I_0$ denotes the set of nonempty finite subsets of $I$, and $x_J=\prod_{j\in J}x_j$.

If $I$ is finite, then $R_I$ is finite and nilpotent (of length $\le 2^{|I|}$ but I'm lazy to get the best bound).

Then $R_I$ is not T-nilpotent when $I$ is infinite. But now let $I$ denote the positive integers, and define $S$ as the quotient modding out by $(x_ix_j)$ whenever $2i<j$. It has the basis $(x_J)$ where $J$ ranges over nonzero finite subset not containing any pair of the form $\{i,j\}$ with $2i<j$.

(a) $S$ is quasinilpotent (with $n=2$ as any X-ring)

(b) $S$ is not nil (because $x_n\dots x_{2n}\neq 0$ for every $n$)

(c) $S$ is T-nilpotent: indeed, consider a sequence $(y_n)$. Write $y_1=\sum_{k=1}^mx_{J_k}$, with $J_k$ nonempty, pairwise distinct and not containing any pair $\{i,j\}$ with $2i<j$; let us show $\prod_{i=1}^p y_k=0$ for large $p$. We can assume $m>0$ since otherwise $y_1=0$. Define $n=\max\bigcup J_k$. Let $\pi$ map $x_J$ to itself if $J\subset\{1,\dots,2n\}$ and map $x_J$ to 0 otherwise, and extend it by linearity; this is a projection, and $y_1y=y_1\pi(y)$ for all $y\in S$. Hence $y_1\dots y_p=y_1\pi(y_2)\dots \pi(y_p)$. Since this product lies in the (finite) subring generated by $x_1,\dots,x_{2n}$, which is nilpotent, it is 0 for large $p$ (say $p\ge 4^n$).

No need to specify the ambient ring since every commutative associative ring $I$ (possibly without unit) is an ideal in a ring (namely $I\oplus\mathbf{Z}$ with multiplicative law $(a_1,n_1)(a_2,n_2)=(a_1a_2+n_1a_2+n_2a_1,n_1n_2)$).

Now call a X-ring a ring (possibly without unit) satisfying the axioms: associative, commutative, $x+x=x^2=0$ for all $x$.

So the free X-ring $R_I$ on the generators $(x_i)_{i\in I}$ has the basis over the field on 2 elements: $(x_J)_{J\in 2^I_0}$, where $2^I_0$ denotes the set of nonempty finite subsets of $I$, and $x_J=\prod_{j\in J}x_j$.

If $I$ is finite, then $R_I$ is finite and nilpotent (of length $\le 2^{|I|}$ but I'm lazy to get the best bound).

Then $R_I$ is not T-nilpotent when $I$ is infinite. But now let $I$ denote the positive integers, and define $S$ as the quotient modding out by $(x_ix_j)$ whenever $2i<j$. It has the basis $(x_J)$ where $J$ ranges over nonzero finite subset not containing any pair of the form $\{i,j\}$ with $2i<j$.

(a) $S$ is quasinilpotent (with $n=2$ as any X-ring)

(b) $S$ is not nilpotent (because $x_n\dots x_{2n}\neq 0$ for every $n$)

(c) $S$ is T-nilpotent: indeed, consider a sequence $(y_n)$. Write $y_1=\sum_{k=1}^mx_{J_k}$, with $J_k$ nonempty, pairwise distinct and not containing any pair $\{i,j\}$ with $2i<j$; let us show $\prod_{i=1}^p y_k=0$ for large $p$. We can assume $m>0$ since otherwise $y_1=0$. Define $n=\max\bigcup J_k$. Let $\pi$ map $x_J$ to itself if $J\subset\{1,\dots,2n\}$ and map $x_J$ to 0 otherwise, and extend it by linearity; this is a projection, and $y_1y=y_1\pi(y)$ for all $y\in S$. Hence $y_1\dots y_p=y_1\pi(y_2)\dots \pi(y_p)$. Since this product lies in the (finite) subring generated by $x_1,\dots,x_{2n}$, which is nilpotent, it is 0 for large $p$ (say $p\ge 4^n$).