How to find t value without data? Consider a simple linear regression model:
$y=\beta_0+\beta_1x+\epsilon,$
where $\epsilon$~$N(0,\sigma^2)$.
The number of observations is 13,
              estimate     standard error      t value     p-value      
$\beta_0$       -0.44           0.2               ?           ?
$\beta_1$        0.31           0.1               ?           ?
R-squared:0.95

T table link is in below.
https://www.statisticshowto.datasciencecentral.com/tables/t-distribution-table/
I don`t know how to get $t$ value without data set.
Question.
How to get t value with limited info?
 A: T-Statistic and p-value: the recipe
You can find the t-statistic by dividing your estimated coefficient by its estimated standard error:
$$t_{\widehat{\beta}} = \frac{\widehat{\beta}}{\widehat{SE_{\beta}}}.$$
Based on this you can derive the corresponding p-value based on the t-distribution. The t-distribution depends on the number of the so-called degrees of freedom. For linear regression the number of degrees of freedom $d$ is the number of observations $n$ minus the number of estimated coefficients $k$, i.e. the number of betas:
$$d = n - k.$$
Then, for $F_d$ being the cumulative distribution function of the t-distribution with $d$ degrees of freedom the p-value can be obtained as
$$ p_{\widehat{\beta}} = (1 - F_d(|t_{\widehat{\beta}}|)\cdot 2.$$
Note that multiplication by 2 is necessary since you are doing usually a two-sided test in linear regression. In other applications where one-sided-tests are used this is not necessary. Moreover, it is necessary to use the absolute value of $t$.
In case the p-value is not relevant for you and you are only testing whether the coefficient is significantly different from zero, it suffices to find the confidence interval for $\beta$ for a certain significance level $\alpha$. Then, the confidence interval is constructed as
$$ \widehat{\beta} \pm \widehat{SE_{\beta}}\cdot t_{d, \alpha /2}$$

Example: doing the math for $\beta_0$
In your example for $\beta_0$ this would be
$$t_{\widehat{\beta_0}} = \frac{-0.44}{0.2} = -2.2$$.
The degrees of freedom $d$ are
$$d = 13 - 2 = 11$$
Now, you get the p-value by computing at first the probability of the cumulative distribution function corresponding to your $t$. You can take a statistical software for this, e.g. R:
pt(2.2, 11)  # 2.2 - Absolute value of t-stats and 11 degrees of freedom
# Result: 0.9749569

Then you get your p-value by subtracting that value from 1 and multiplying it by 2. 
$$ p_{\widehat{\beta_0}} = (1-0.9749569)\cdot2 = 0.0500862. $$
For the confidence interval let's say you set $\alpha = 0.05$, i.e. you construct the 95% confidence interval. You can take the corresponding $t_{d, \alpha /2}$ from the link to the table you posted in your question: 
$$t_{d, \alpha /2} = t_{11, 0.025} = 2.201.$$
Then, the 95% confidence interval is 
$$ -0.44 \pm 0.2 \cdot 2.201 = [-0.8802, 0.0002]. $$
Based on the p-value and confidence interval you can conclude that $\beta_0$ is not significantly different from 0 for a significance level of 5%. This can be seen because the p-value is greater than 0.05 and the confidence interval contains zero, i.e. the lower bound is negative and the upper bound positive.
Now you see you can easily compute everything without knowing the data. So, try it yourself for $\beta_1$!
