Proof: Proceed by induction
Base case $P(t=1)=n_1$, which is even since $n$ is an even integer. True. Assume $P(t)$, i.e. if $n_1,n_2,\cdots n_t$ are even, then $n_1+n_2+...+n_t$ is even, is true.
Inductive step: $P(t+1)=n_1+n_2+...n_t+n_{t+1}$. Since $n_1+n_2+...+n_t$ is assumed to be even and $n_{t+1}$ is even, an even $+$ even gives an even number. Therefore $P(t+1)$ is shown to be true by induction.
Please let me know if this proof is satisfactory as I am relatively new to induction proofs.
You have the right idea, but some of your notation and phrasing is off.
First, note that a statement like "$P(t=1)=n_1$" doesn't make sense, for several reasons. First, you've never defined what $P(t)$ means. Second, even if you had defined it, $P(t=1)$ would not make sense; instead you would just write $P(1)$. Third, the usual thing to call "$P(t)$" here would be the statement that if $n_1,\dots,n_t$ are even integers, then so is their sum. So $P(t)$ (or $P(1)$) is an entire statement, not just a number. It doesn't make sense to say $P(1)=n_1$.
What you can instead say for your base case (after defining what $P(t)$ means) is something like:
Note how I clearly introduced the variable $n_1$ with "suppose $n_1$ is an even integer" before proceeding to prove anything. This is in general a good practice when proving a "for all" statement: you first start by explicitly mentioning that you are considering a particular instance of the statement.
There are similar issues with your phrasing in the inductive step. Also, the induction hypothesis (where you assume that $P(t)$ is true) should be the start of the inductive step, rather than the end of the base case.