Is every infinite sequence of integers in $ T $ ?

Yup $T$ contains every possible infinite sequence of integers.

Suppose there are primes $p_1, p_2, \dots $ such that each $a_k$ in the sequence is divisible by $p_1p_2 \dots p_k$, then $f(a_1, \dots, a_{k-1}, a_k, a_{k+1}, \dots) = f(0, \dots, 0, a_k, a_{k+1}, \dots)$ is divisible by $p_1p_2 \dots p_k$. Thus for such sequences $x \in T$, $f(x) = 0$. To finish the proof, note that every sequence in $T$ can be written as the sum of two such sequences (with distinct set of associated prime divisors).

Let me rewrite angiland's proof more explicitly. Note that for any integers $a_1, a_2, \ldots$, \[ 2^n \mid f(0, \ldots, 0, 2^n a_n, 2^{n+1} a_{n+1}, \ldots) = f(2 a_1, 4 a_2, 8 a_3, \ldots) \]for all $n > 0$ and hence $f(2 a_1, 4 a_2, \ldots) = 0$. Likewise $f(3 b_1, 9 b_2, \ldots) = 0$ for arbitrary integers $b_1, b_2, \ldots$. For each $(c_1, c_2, \ldots) \in T$, there exist integers $a_k, b_k \in \mathbb{Z}$ such that $2^k a_k + 3^k b_k = c_k$. Then \[ f(c_1, c_2, \ldots) = f(2 a_1, 4 a_2, \ldots) + f(3 b_1, 9 b_2, \ldots) = 0 + 0 = 0. \]

Thanks I hate it. We are given that for all $x$ with finitely many nonzero elements, $f(x)=0$. (Rigorously, to show $f(x)=0$ for $x=(a_1,a_2,a_3,\ldots,a_n,0,0,\ldots)$, let $x_i$ be the sequence with $i$th term $1$ and all other terms $0$, and take $\textstyle f(x)=\sum_{i=1}^na_if(x_i)$.) Consequently for any sequence $x=(a_1,a_2,a_3,\ldots)$, we can subtract off $(a_1,a_2,a_3,\ldots,a_n,0,0,\ldots)$ and assume the first $n$ elements of the sequence are $0$. Note that for all positive integers $n$ and sequences of integers $(b_1,b_2,b_3,\ldots)$ and $(c_1,c_2,c_3,\ldots)$, we have \begin{align*} f(2b_1,2^2b_2,2^3b_3,\ldots)&=f(0,0,\ldots,0,2^nb_n,2^{n+1}b_{n+1},\ldots)\\ &=2^nf(0,0,\ldots,0,b_n,2b_{n+1},\ldots)\\ &\equiv0\pmod{2^n}, \end{align*}and similarly \[f(3c_1,3^2c_2,3^3c_3,\ldots)\equiv0\pmod{3^n}.\]It follows that they are both $0$. For any sequence $(a_1,a_2,a_3,\ldots)$, by B\'ezout's lemma we can find sequences of integers $(b_1,b_2,b_3,\ldots)$ and $(c_1,c_2,c_3,\ldots)$ such that $a_n=2^nb_n+3^nc_n$ for all positive integers $n$. Thus \[f(a_1,a_2,a_3,\ldots)=f(2b_1,2^2b_2,2^3b_3,\ldots)+f(3c_1,3^2c_2,3^3c_3,\ldots)=0,\]and we are done.

The result here is intimately related to the properties of the Baer-Specker group—see also Example 3.5. here and this MathOverflow question—as is in particular implied by the inclusions: \begin{align*}0\to\coprod_{i\colon\mathbb N}\mathbb Z\overset{i}{\longrightarrow}\prod_{i\colon\mathbb N}\mathbb Z & \ \ \ \ \ \text{(of }\mathbb Z\text{-modules)} \\ 0\to\left[\prod_{i\colon\mathbb N}\mathbb Z,\mathbb Z\right]_{\mathbb Z}\overset{\left[i,\text{id}_{\mathbb Z}\right]}{\longrightarrow}\left[\coprod_{i\colon\mathbb N}\mathbb Z,\mathbb Z\right]_{\mathbb Z} & \ \ \ \ \ \text{(of internal homs of }\mathbb Z\text{-modules)}\end{align*}

The devil weeps. Claim: If $z \in {\mathbb Z}^\infty$ only has finitely many zero terms then it is $f(z) = 0$. Proof. Follows inductively. $\blacksquare$ Claim: If all terms in $z = (z_1, z_2, \dots)$ are divisible by $k$, then $k$ also divides $f(k)$. Proof. Since $f$ is additive, we can factorize out $k$ to get that \[ f(z) = k \cdot f\left(\left(\frac{z_1}{k}, \frac{z_2}{k}, \dots\right)\right) \]$\blacksquare$ Claim: For any $a = (a_1, a_2, \dots) \in {\mathbb Z}$ and integer $n \ge 2$, then the sequence $t = (a_1, a_2n, a_3n^2, \dots)$ is mapped to zero by $f$. Proof. This is equivalent to $n^e \mid g(a, n)$ for any positive integer $e$, which follows by truncating the first $e$ terms. $\blacksquare$ Now, let $a = (a_1, a_2, \dots)$ be an arbitrary sequence. Define $b = (b_1, 2 \cdot b_2, 2^2 \cdot b_3, \dots)$ and $c = (c_1, 3 \cdot c_2, 3^2 \cdot c_3, \dots)$ such that $2^i \cdot b_i - 3^i \cdot c_i = a_i$, which occurs for some integers $b_i$ and $c_i$ by Bezout's. As such, \[ f(a) = f(b - c) = f(b) - f(c) = 0. \]

Great problem Claim: For all $x, y\in T$, if $x$ and $y$ differ in finitley many positions then $f(x)=f(y)$ Claim: If $a$ divides all the elements of some $z\in T$, we have that $a$ must also divide $f(z)$ Proof. Say $z=(az_1, az_2, \dots)$. Due to the additive structure of $f$, we can factor out $a$ to get that $f(z)=af(z_1,z_2\dots)$ as desired $\blacksquare$ Now comes the main claim Claim: If $(z_0,z_1,z_2\cdots)\in T$, then for all integers $n$ we have that $f(k)=0$ where $k=(z_0, n\cdot z_1, n^2\cdot z_2..$) Proof. Due to the first and second lemma we can easily check that $f(k)$ has arbritrarily large divisors and thus the conclusion follows $\blacksquare$ Now comes the final blow. For any $a\in T$, by bezouts take a $b,c$ such that $b,c$ and the $i$th term of $b$ and $c$ respectivley are divisible by $2^i$ and $3^i$ and $a=b-c$. Thus we get that $$f(a)=f(b-c)=f(b)-f(c)=0$$as desired. Note: The notation is almost the exact same as above and this was purely coincidental