The answer is in part I of Linear Operators by Dunford and Schwartz, Theorem VI.1.4. For any Banach spaces $X$ and $Y$ linear functionals on $\mathcal{B}(X,Y)$ continuous in strong and weak operator topologies are the same. It follows from the proof that they are of the form $F(A)=\sum_{i=1}^n\varphi_i(Ax_i)$ for $x_i\in X$ and $\varphi_i\in Y^*$. In other words, the dual space can be identified with the algebraic tensor product $X\otimes Y^*$, or equivalently the space of finite rank operators $\mathcal{F}(Y,X)$.
For the ultraweak and the ultrastrong topologies on $\mathcal{B}(H)$, where $H$ is a Hilbert space, the answer is in Von Neumann Algebras by Dixmier, Lemma I.3.2. Again, the dual spaces coincide and can be identified with the projective tensor product $H\otimes_\pi H$, or the space of the trace class operators $\mathcal{L}_1(H)$. It also happens to coincide with the norm closure of $\mathcal{F}(H)$ in the Banach dual $\mathcal{B}(H)^*$, and is a Banach predual $\mathcal{B}(H)_*$ of $\mathcal{B}(H)$. Moreover, the ultraweak topology on $\mathcal{B}(H)$ coincides with the weak* topology generated by this predual, they are both $\sigma\big(\mathcal{B}(H),H\otimes_\pi H\big)$.
It doesn't appear that the ultraweak or the ultrastrong topologies were studied much beyond Hilbert spaces, I couldn't even find any established definitions for them. Since the weak operator topology coincides with $\sigma\big(\mathcal{B}(X,Y), X\otimes Y^*\big)$, by analogy to Hilbert spaces a natural candidate for the ultraweak is $\sigma\big(\mathcal{B}(X,Y), X\otimes_\pi Y^*\big)$. However, according to Introduction to Tensor Products of Banach Spaces by Ryan, Section 2.2 a predual to $\mathcal{B}(X,Y)$ is $X\otimes_\pi Y_*$, where $Y_*$ is a predual to $Y$. So the ultraweak so defined does not coincide with the weak* unless $Y$ is reflexive. And even if $Y$ is reflexive it is not clear what a natural candidate for the ultrastrong is or how the duals are related.
Note that if $x=(x_n)\in\ell^1$, then $x\in\ell^\infty$ as well. Now
$$\|T_nx-x\|_{\ell^\infty}=\|(0,0,,\dots,0,x_n-x_{n+1},x_n-x_{n+2},x_n-x_{n+3},\dots)\|_\infty=\sup_{k\geq1}|x_n-x_{n+k}|$$
But since $x\in\ell^1$ we have that $\{x_n\}$ is a Cauchy sequence (since it converges (to $0$)). So if $\varepsilon>0$ there exists $N\geq1$ so that $|x_n-x_m|<\varepsilon$. Then, for $n\geq N$ we have that
$$\|T_nx-x\|_\infty=\sup_{k\geq1}|x_n-x_{n+k}|\leq\varepsilon$$
which shows that $T_nx\to x$ in $\ell^\infty$. In other words, since this is true for all $x$, we have that $T_n\to I$ strongly, where $I:\ell^\infty\to\ell^\infty$ is the identity operator.
comment: Note that the same would be true if we "enlarged" the domain of each $T_n$ and used the space $c$ of convergent sequences with supremum norm.
Edit
Here are some details about checking whether $T_n$ converges in norm, as OP seems to have trouble with this part.
First, if $T_n$ converges in norm to something, it has to be $I$, because if $T_n$ converges in norm to $S$, then $T_n$ converges strongly to $S$ and therefore $S=I$ (strong operator limits are unique).
First let's compute the norm of $T_n$, but this is not needed to check whether $T_n\to I$ in norm; we should be computing $\|T_n-I\|$ for this, but let's start with that. Note that $n$ is fixed now! We have that $$\|T_n\|=\sup_{\|x\|_{\ell^1}=1}\|T_nx\|_{\ell^\infty}=\sup_{\|x\|=1}\|(x_1,\dots,x_n,x_n,x_n,\dots)\|_{\ell^\infty}=\sup_{\|x\|=1}\bigg\{\max_{1\leq j\leq n}|x_j|\bigg\}\leq\sup_{\|x\|=1}\bigg\{\sum_{j=1}^\infty|x_j|\bigg\}=1 $$
So $\|T_n\|\leq1$. On the other hand, if $x=(1,0,0,\dots)$ then $\|x\|_{\ell^1}=1$ and $T_nx=(1,0,0,\dots,)$ and $\|T_nx\|_{\ell^\infty}=1$, so
$$\|T_n\|=\sup_{\|y\|_{\ell^1}}\|T_n(y)\|\geq\|T_n(x)\|=1$$
and this shows that $\|T_n\|=1$ for all $n$.
Now let's compute in the same fashion the norm of $T_n-I$, again $n$ is fixed. We do not really need to do an exact calculation, we only need to know if $\|T_n-I\|$ converges to $0$ or not. By our earlier computations before the edit,
$$\|T_n-I\|=\sup_{\|x\|=1}\|T_nx-x\|=\sup_{\|x\|=1}\sup_{k\geq1}|x_n-x_{n+k}|$$
Now consider $x=(0,\dots,0,0,1,0,\dots)$ where $1$ appears in the $n+1$ position. Then, $\sup_{k\geq1}|x_n-x_{n+k}|=sup_{k\geq1}|x_{n+k}|=|x_{n+1}|=1$. Also note that $\|x\|_{\ell^1}=1$, so,
$$\|T_n-I\|\geq\|T_nx-x\|_{\ell^\infty}=1$$
This shows that the sequence $\{\|T_n-I\|\}_{n=1}^\infty$ is ALWAYS greater than or equal to $1$, so it cannot converge to $0$.
Best Answer
We want to show that a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 1 if and only if a net $T_{\lambda}$ converges to $T$ with respect to $\textbf{def}$ 2.
Suppose that $T_{\lambda}\to T$ with respect to $\textbf{def}$ 1, then we know that $T_{\lambda}x\to Tx$ for each $x\in X$ meaning that $$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T).$$ Putting this in context, for each $F_x\in \{F_x:\beta(X,Y)\to Y \text{ such that }F_x(T)=Tx, T\in\beta(X,Y)\}$ we have $F_x(T_{\lambda})\to F_x(T)$. So $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2.
On the other hand, if $T_{\lambda}\to T$ with respect to $\textbf{def}$ 2, then we know that $$F_x(T_{\lambda})=T_{\lambda}x\to Tx=F_x(T),$$ for each $x\in X$ and as a result $T_{\lambda}x\to Tx$ for each $x\in X$ and we have convergence with $\textbf{def}$ 1. It is simply unpacking the definitions.